Mole concept numerical ..........

Source : Banglore institiute of coaching ii PUC coaching

1. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S S OL U TIO N S S o me m o r e A d d i t i on al P r ob l ems on S o l u t i ons ( S o l v e d ) 1 . W h a t is t h e m o l a r ity o f a n a q u e ou s s o l u ti o n o f e t h y l a lc oho l ( C H 3 CH 2 OH, m o l ar m a ss = 4 6 .0 g / m o l) w h ich c on t a i n s 5 .0 g o f a lc oho l in 5 m L o f t h e s o l u t i on ? S olu t ion. 5 × 1000 M a ss o f e t h y l a lc oho l/ lit r e = = 100g 500  M ass i n g / li tr e  10 N o . o f m o l e s o f a lc o ho l / lit r e = = = = 2.18 M o l ar m ass i n g /m ol 4 6 T h u s, t h e s o l u ti o n c o n t a i n s 2 . 1 8 m o l e s o f e t h y l a lc oho l pe r lit r e o f t h e s o l u t i on . So , t h e m o l a r ity o f t h e s o l u ti o n is 2 . 1 8 m o l / L. 2 . C a lc u l a te t h e no rma lit y · an d m o l a r ity o f H 2 S0 4 in a s o l u ti o n c on t a i n i n g 9 .8 g o f H 2 S O 4 pe r d m 3 o f t h e s o l u ti o n. S olu t ion : M a ss o f H 2 S0 4 pe r lit r e = 9 .8 g M o l a r m a ss o f H 2 S O 4 = 2 x ( 1 ) + 3 2 + ( 4 x 16 ) = 2 + 3 2 + 6 4 = 9 8 g m o l - 1 98 T h e re f o re, N u m b e r o f m o l e s o f H 2 S0 4 p e r lit r e o f s o l u ti o n = = .1 98 M o l a r ity o f H 2 S0 4 i n s o l u ti on = .1 m o ll 1 - 1 H en c e, A s t he r e a r e t w o e q u iv a l ent s pe r m o le o f H 2 S O 4 , t he re f o r e N o r m a lity o f H 2 S0 4 'in s o l u ti on = .1 x 2 = . 2 . eq u iv L - 1 3 . 2 . 8 2 g o f g l u c o se (m o l a r m a ss 1 8 g m o l -1 ) a r e d iss o l v e d in 3 g o f w ate r C a lc u l a t e. ( i) m o l a lity o f t h e s o l u t i o n ( ii) m o l e f r a cti on s o f , ( a ) g l u c o se ( b ) w a t e r . S olu t ion. M a ss o f g l u c o s e , w 8 = 2 : 8 2 g M g = 18 g m o t - 1 M o l a r m a ss o f g l u c o s e , M a ss o f w a t e r , w w = 3 g BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 1

2. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S M o l a r m a ss o f w a t e r , M g = 1 8 g m o l - 1 ( i) F r o m t h e d e f i n iti on , m o l a lity is t h e n u m b e r o f m o l e s pe r 1 00 g o f t h e s o l v en t. T h e n, 2.28 N u m b e r o f m o l e s o f g l u c o se m 3 g w a t er = m o l = . . 1 5 7 m ol 180 0.0157 × 100 = . . 5 2 2 ' m o l k g - 1 So , M o l a lity o f g l u c o s e = 30 ( ii) F r o m abo v e , t h e n u m be r o f m o l e s o f g l u c o se ( n g ) a n d . w a t e r ( n w ) in t h e s o l u ti o n a r e g i v e n b y , M a ss o f g l u c o se 282 n = = m o l = · 1 5 7 m ol M o l a r m a ss o f g l u c o se 180 g M a ss o f w ater 30 n w  = m o l = 1.667 m ol M o l a r m a ss o f w ater 18 n g 0.0157 T h e re f o re, M o l e f r a cti o n o f g l u c o s e = = = 0.01 n g + n w 0.015 7 + 1.667 n w 1.667 M o l e f r a cti o n o f w a t e r = = = 0.99 n g + n w 1.66 7 + 0.0157 4 . W h a t is t h e m o l a lity o f t h e a qu eou s s o l u ti o n o f m e t h y l a lc oho l ( C H 3 OH. m o l a r m a ss 32 .0 g / m o l) w h ich c on t a i n s 6 4 g o f m e t h y l a lc oho l in 2 g o f w a t e r ? S olu t ion: M a ss o f m e t h y l a lc oh o l in 2 g o f w a t e r = 6 4 g 6 4 × 1000 = 320 g 1 T h en , M a ss o f m e t h y l a lc oh o l in 200 320g So , N o . o f m o l e s o f a lc o ho l pe r 1 00 g o f w a t e r = 1 m ol 32g / m ol T h u s, I kg o f t h e s o l u ti o n c o n t a i n s 1 m o l e o f m e t h y l1 a lc o h o l. S o , t h e m o l a lity o f t h e s o l u ti o n is 1 m o l /k g. BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 2

3. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 5 . A n a g u e ou s o f s o l u t i o n o f a . d i b a sic a cid ( m o l a r m a ss =1 1 8 g / m o l ) c on t a i n i n g 17 . 7 g o f t h e a c i d p e r lit r e o f s o l u ti o n ha s a d en sit y , I. 00 7 7 g / m L . E x p r e ss t h e c o n c e n t r a ti o n o f t h e s o l u ti o n in a s m an y a s y o u c an S olu t ion. M a ss o f s o l u t e, M o l a r m a s s " o f t h e s o l u t e, w M = 17 .7 g /L = 11 8 g m o l - 1 M a ss o f s o l ut e ( g/L ) w So 'No · o f m o l e s o f s o l u t e d iss o l v ed = = m o l / L -1 M o l a r m a ss ( g m p l ) M 17. 7 T h en M a ss o f t h e s o l v en t, ( w ate r ) Vo l u m e o f t h e s o l u ti on D en sity o f s o l u ti o n M a ss o f s o l v en t ( w a t e r ) = M o l /L = . 1 5 m o l /L 118 = 100 m L · = 1 . 7 7 g / mL = Vo l u m e x D en sity = 10 m L x 1 . 077 g / m L = 100 7 .7 g = M a ss o f s o l u ti o n - M a ss o f s o l u t e = ( 1007 . 7 - 1 7 .7 g ) = 9 9 g So , T h en, M aa s o f t h e s o l v en t, (w a t e r ) M o l e s o f s o l ut e 9 9 N o . o f m o l e s o f s o l v e n t = = = 55. M o l a r m a ss i n g / m o l 18 T o t a l n u m be r o f m o l e s in s o l u ti o n = M o l e s o f s o l u te + Mo l e s o f s o l v en t = . 1 5 +5 5 = 5 5 . 15 S i n c e , t h e a cid is d i ba s ic, hen c e its E q . m a ss = M o l a r m a s s/ 2 = 5 9 g e q u iv - 1 F r o m t h e a bo v e r e s u lts t h e c on c en t r at i o n o f s o l u ti o n c a n b e c a l c u l a t e d in v a r i ous un its a s f o ll o w s: ( i) M o l a r i t y : A l r ead y c a lc u l a t e d a b o v e : M o l a r i t y = . 1 5 m o l/L ( ii) M o l a li t y : F r o m a b o v e, 1000g 1 M a ss o f s o l u te ( a ci d ) p e r 10 g o f s o l v en t =1 7 .7 × × = 0.151 m o l / k g 99 g 1 18 M a ss i n g pe r 100 g s o l v ent N o . o f m o l e s o f t h e s o l u t e / 1 g o f s o l v en t = M o l a r m a ss ( g / m o l ) 1000 1 = 17 .7 × × = 0.15 1 m o l / k g 990 118 So , M o l a lity = . 15 1 m o l/ k g BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 3

4. ADD I T IO N AL PR OB L E MS ON S O L U T I O N S n a c i d 0.15 0.15 ( iii) M o l e - f r a cti on ( X ) = = = = 0.027 n n a c i d + w ater 0.1 5 + 55 55.15 6 . H o w m u ch s od i u m c h l o r i d e b e d is s o l v e d t o m a ke l lit r e o f . 1 F s o l u ti o n ? S olu t ion : . N a Cl. T h e re f o re, So , Sod i u m c h l o r i d e is a n i on ic .c o m p ou n d a n d is re p r e s e n t e d b y t h e f o rm u la its gr a m - f o rm u la m a ss is, 2 3 g + 3 5 .5 g 1 lit r e o f 1 F N a Cl s o l u t i on , o n e r e q u i r es 1 lit r e o f .1 F N a CI s o l u ti on , o n e . r e q u i r es = 58 . 5 , g = 58 .5 g , o f N a Cl = 58 .5 x O. 1 g = 5 . 8 5 g T h u s, 5 . 8 5 g o f N a Cl s hou ld b e d iss o l v e d to m a ke u p 1 lit r e o f s o l u t i on. 7 . H o w m an y gr a m s o f K CI w ou l d b e r e q u i r e d t o p re pa r e 1 lit r e o f . 1 M s o l u ti o n ? A t o m ic m a s s e s a r e ; K = 3 9 u , Cl - = 3 5 .5 u S olu t ion: P o t a ssi u m c h l o r i d e ( K Cl) i s a n i on i c c o m po u n d w ith a m o l a r m a ss o f 3 9 + 3 5 .5 = 74 . 5 g m o l - 1 T h e re f o r e, F o r 1 lit r e o f 1 M K Cl s o l u ti o n , o n e . r e q u i r es 74 .5 g o f K Cl 74. 5 × 0. 1 F o r 1 lit r e o f . 1 M K Cl s o l u ti o n , o n e r e q u i r es = 7.45g 1 T h u s, o n e r e q u i r e s 7 . 4 5 g o f K Cl t o p re pa r e 1 lit r e o f its .1 M s o l u ti o n 8 . 2 . 4 6 g o f s o d i u m h y d r o x i d e ( m o l a r m a ss = 4 g / m o l) a r e d is s o l v e d in w a t e r a n d t he s o l u ti o n is m ad e to . 1 m L in a v o l u m e t r ic f l a sk. C a lc u l a t e t h e m o l a r ity o f t h e s o l u ti o n . S olu t ion : M a ss o f s o d i u m h y d ro x i de = 2 . 4 '6 g M o l a r m a ss o f s od i u m h y d r o x i d e ·= 4 g m o ll - 1 Vo l u m e o f t h e · s o l u ti o n M o l a r ity o f t h e s o l u ti on = 10 m L =? M o l a r ity o f a s o l u ti o n i s g i v e n b y t h e n u m be r o f m o l e s o f s o l u t e p r e s e n t in 10 m L of t h e s o l u ti o n , i. e ., N o . o f m o l e s o f s o l ut e x 100 m l L - 1 M o l a r ity = Vo l u m e o f th e s o l u t i o n i n m L BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 4

5. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S (M a ss o f th e s o l ut e I M o l a r m a ss o f th e s o l ute) x 100 m o l L - 1 = Vo l u m e o f th e s o l ut i o n i n m L 2.46 / 4 ) × 1000 2.46 -1 × 10 m o l L - = 0.615 m o l L = m o l L - 1 = 100 40 So , t h e m o l a lity o f t h e s o l u ti o n is 6 1 5 m o l L - 1 9 . C a lc u l a te t h e m o l a lity o f a 1 M s o l u ti o n o f s od i u m n it r a t e . T h e de n si t y o f t h e s o l u ti on is, t h e s o l u ti o n is 1 . 2 5 g c m - 3 S olu t ion: M o l a r m a ss o f s o d i u m n it r a t e, ( N a N 3 ) = ( 2 3 + 1 4 + 4 8 ) g / m o l = 8 S g / m o l M a ss o f 1d m 3 ( o r 1 lit r e ) o f t h e s o l u ti o n = Vo l u m e x D en sity T h e re f o r e, = 100 c m 3 x 1 . 2 5 g / c m 3 = 1 2 5 g T h e re f o re, M a ss o f w a t e r c o n t a i n i n g 8 .5 g . o f N a N 3 = ( 12 5 - 85 ) g = 1 1 6 5 g = 1 . 1 6 5 kg So , 1 m ol -1 M o l a lity ( m ) o f t h e s o l u ti on. = = 0.8 6 m o l k g 1.165 k g 10 . . 7 5 .g o f s od i u m b i c a r bo n a te ( N a HC 3 ) a r e d iss o l v e d i n 2 5 m l o f a s o l u ti on. C a lc u l a te its, ( i) no rm a li t y , ( i i ) m o l a r ity S olu t ion . M a ss o f s o d i u m b ic a r b ona t e d iss o l v e d Vo l u m e o f s o l u ti on = . 75g = 25 m L L - M o l a r m a ss o f s od i u m b ic a r bo n a te . = ( 23 + 1 + 12 +~ 8 ) g m o l - 1 = . 6 15 1 g m o t' S i n c e , on e m o l e c u le o f N a HCO 3 c on t a i n s o n l y on e c a ti o n i c c ha rg e ( o n N a +) , h en ce M o l a r m a ss 84 = g equ i v -1 E q u i v a l en t m a ss, ( E )= 1 1 A s pe r d e f i n iti o n, (M a ss o f N a HC O 3 / M o l a r m a ss o f N a HC 3 ) M o l a r ity ( M ) = x 1000 Vo l u m e o f s o l ut i o n i n m L -1 0.75g/8 4 m ol 0.75/84 m ol -1 = 0.0357 m o l L = × 1000 m L ×  250/100  L - 1 25 m L BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 5

6. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S (M a ss o f N a HC O 3 / Equ i v . m a ss o f N a HC 3 ) × 100 M L / L N o r m a lity ( N) = Vo l u m e o f s o l ut i o n i n m L 11 . C a l c u l a te t h e m o l e f r a ct i o n o f w a t e r in a m i x t u r e o f 1 2 g w a t e r . 1 8 g a c e t i c a cid a nd 9 2 g e t h y l a lc oho l. S olu t ion : F o ll o w i n g t h e p r o c e d u r e o f t h e p r e v i ou s p ro b l e m , o n e c a n w r ite M a ss o f w ater 12 n ( H 2 O) = = = 0.67 M o l a r m a ss M a ss o f ethanol 18 92 n ( C 2 H 5 OH) = = = 2 . M o l a r m a ss 46 M a ss o f a c e t i c a c i d 108 n ( CH 3 COOH) = = = 1.80 M o l a r m a ss 60 So , T o t a l n u m be r o f m o l e s in t h e s o l u ti o n , n t o t a t = . 6 7 + 2 . + 1 . 8 = 4 . 47 n  H 2 O  0.67 T h e re f o re, X w ate r = = = 0.15 n t o t a l 4.47 n  C 2 H 5 O H  2.00 X ethano l = = = 0.45 n t o t a l 4.47 n  C H 3 c oo H  1.80 X a c e t i c a c i d = = = 0.40 n t o t a l 4.47 12 . W ha t is t h e m o l a li t y o f am m o n ia in a s o l u ti o n c o n t a i n i n g . 8 5 g N H 3 in a 1 c m 3 o f a li q u id o f d e n sity . 8 5 g c m - 3 ? S olu t ion: M a ss o f am m o n ia w 2 = . 8 5 g M 2 = 1 7 g m o l - 1 w 1 = 10 c m 3  . 8 5 g c m -3 = . 8 5 g M o l a r m a ss o f a m m o n i a, M a ss o f li q u id ( s o l v ent ) , BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 6

7. ADD I T IO N AL PR OB L E MS ON S O L U T I O N S n NH w /M -1 3 2 2 = m o l k g So , M o l a lity o f am m o n ia = M a ss o f s o l v en t i n k g 85/1000 0.85/17  85/100  m o l k g - 1 = . 5 9 m o l k g - 1 o r, M o l a lity o f am m o n ia = 13 . A n a qu eou s s o l u ti o n o f a d i ba sic a cid (C 2 H 2 O 4 . 2 H 2 O) c on t a i n s 1 . 2 6 g o f t h e s o l u te pe r lit r e o f t h e s o l u t i o n C a lc u l a te t h e no r m a lity en d m o l a r ity o f t h e s o l u ti on S olu t ion: M a ss o f t h e a cid W = 1 . 2 6 g /L =    2 × 1 2  +  2 × 1  +  4 × 1 6  + 2  2 + 1 6  = 2 4 + 2 + 6 4 + 3 6 = 12 6 g m ol M o l a r m a ss M o f d i ba s ic a cid -1 So, M o l a r m a ss 126 -1 E q u i v a l en t m a ss o f t h e a c i d = = = 6 3 g equ i v 2 2 A cc o r d i n g to t h e d e f i n i t i on, M a ss o f s o l ut e i n g pe r L 1.2 6 g / L -1 M o l a r ity = = = 0.01 m o l L M o l a r m a ss 12 6 g/ m ol M a ss o f s o l ut e i n g pe r L 1.2 6 g / L -1 and , N o r m a lity = = - 1 = 0.01 m o l L Equ i v a l en t m a ss 6 3 g equ i v 14 . W ha t v o l u m e o f 9 5 % s u l p h u r ic a cid ( d e n s it y = 1 . 8 5 g /c m 3 ) an d w ha t m a ss o f w a t er m u st to t a k e n to p re p ar e 10 c m 3 o f 1 5 % s o l u ti o n o f s u l phu r ic a cid ( den sit y ) = 1 . 1 g / c m 3 )? S olu t ion: = 10 c m 2 = 1 . 1 g / c m 3 Vo l u m e o f t h e s o l u ti on D en sity o f t h e s o l u t i on T h e re f o r e M a ss o f 10 c m 3 o f s o l u ti o n = 10 x 1 . 1 g = 11 g T h e g i v e n s o l u ti on s is 15 %. T h is m ean s t h a t 10 g o f s o l u ti o n c on t a i n s 1 5 g o f H 2 S O 4 BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 7

8. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 15g T h en , m a ss H 2 S O 4 in 11 g (= 10 c m 3 ) o f s o l u ti o n = × 110 g = 1 6 .5g 100g an d M a ss o f w a t e r in 11 g (= 100 c m 3 ) o f s o l u ti o n = =( 110 - 1 6 . 5 ) g = 93 . 5 g So , to o b t a in 1 c m 3 o f 1 5 % s o l u ti o n a ci d , w e r e q u i r e M a ss o f w a t er = 93 .5 g M a ss o f H 2 S O 4 ( 100 % pu r e) = 16 .5 g S i n c e , t h e g i v e n s u l ph u ri c a cid is 9 5 % p u r e , hen c e 10 × 16.5g M a ss o f H 2 S O 4 ( 95 %) r e q u i r e d = = 17.37g 95 D en sity o f 95 % H 2 S O 4 = 1 . 8 5 c m - 3 15 . S e a w a t e r c on t a i n s 6 x 10 - 3 g o f d is s o l v e d o x y g e n in on e lit r e E x p r e ss t h e c on c e n t r a ti o n o f o x y g e n in s e a w a t e r pe r m illi o n ( p p m ) u n its D en sity o f s ea w a t er = 1 . 3 g / m L. S olu t ion: M a ss o f 1 lit t r e o f s e a w a t e r = 1 m L x 1 . 3 g / m L =1 . 3 g M a ss o f d iss o l v e d o x y g e n i n s e a w a t e r pe r li t r e = 6 x 10 - 3 g 6 × 1 -3  10 6 p p m = 5 . 8 p pm So , C on c e n t r a ti o n o f d iss o l v e d o x y g e n = 103 g 16 . C a l c u l a te t h e m o l a l ity o f a s a lt ( m o l a r m a s s = 13 8 g m o l -1 ) s o l u ti o n o b t a i n e d by d iss o l v i n g 2 .5 g in on e l i t r e o f t h e s o l u ti o n . D e n sity o f s o l u ti o n is . 8 5 g c m - 3 S olu t ion: M a ss o f t h e s a lt d iss o l v ed = 2 .5 g = 100 c m 3 = . 8 5 g c m 3 = Vo l u m e x D en sit y = ( 1 00 x . 85 ) g = 85 g Vo l u m e o f t h e s o l u ti on D en sity o f t h e s o l u t i on So , M a ss o f 10 c m 3 o f s o l u ti on T h e re f o r e M a ss o f s o l v en t = (8 5 0- 2 . 5 ) = 8 4 7 .5 g T h u s 84 7 .5 g o f s o l v en t c on t a i n s 2 .5 g o f s o l u t e BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 8

9. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 2 .5 O r, 847 . 5 g o f s o l v en t c o n t a i ns m o l o f s o l u te 1.38 2 .5 1 1 g s o l v en t c o n t a i n s = × m ol 1.3 8 847. 5 1000 2. 5 10 g o f s o l v en t c on t a i n s = × = 0.021 4 m o l 847 5 138 T h e re f o re , m o l a lity o f t h e s o l u ti on s is . 2 1 4 m o l k g - 1 17 . C a l c u l a te t h e m a ss o f t h e s o l u te p r e s e n t i n t h e f o ll o w i n g s o l u ti o n s: ( a ) l L o f N / 1 N a 2 CO 3 s o l u ti on ( c) 10 m L o f .5 M H 2 S O 4 ( b ) 2 L o f N / 1 HCI s o l u ti on ( d ) 25 m L o f N / 1 o x a lic a cid S olu t ion . ( a ) P r e p a r a t ion of 1 L of N/ 1 s olu t ion s : M o l a r m a ss o f N a 2 C O 3 = ( 2  23 ) + 1 2 + ( 3  16 ) = 1 6 g m o l - 1 W e k no w , t ha t t h e gr a m e q u i v a l en t m a ss o f N a 2 C o 3 is ha l f o f t h e m o l a r m a ss H en c e, 106 = 5 3 g e q u i v - 1 E q u i v a l en t m a ss o f N a 2 CO 3 = = 2 T h u s 1 L o f 1 N N a 2 CO 3 s o l u ti o n s h ou ld c o n t a in 5 3 g o f N a 2 CO 3 53 T h en , 1 L o f N/ 1 N a 2 CO 3 s o l u ti o n w ou ld c o n t a in g o f N a 2 C o 3 = 5 . 3 g N a 2 CO 3 = 5 .3 10 g N a 2 CO 3 ( b) P r e p a r a t ion of 2 L of N/ 1 HCI s olu t ion: E q u i v a l en t m a ss o f H C I = ( 1 + 35 . 5 ) = 3 6 .5 g eq u iv - 1 So , 1 L o f 1 N HCl s o l u ti o n s hou l d c o n t a in 36 . 5 g o f H C I . T h en, BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 9

10. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 36. 5 × 2 73 2 . L o f N / 1 HCl s o l u ti o n w ou ld c o n t a in = g o f HCl = g HCl = 7 .3 g HCl 10 10 ( c ) P r e p a r a t ion of 10 mL of M H 2 S O 4 M o l a r m a ss o f H 2 S O 4 c on t a i n s 9 8 g H 2 S O 4 Or 10 m L o f 1 M H 2 S O 4 c on t a in 9 8 g H 2 S O 4 98 0. 5 T h en , 10 m L o f .5 M H 2 S O 4 w ou ld c on t a i n = × 10 × g o f H 2 S O 4 1000 1 = 4 .9 g o f H 2 S O 4 ( d ) P r e p a r a t ion of 2 5 mL of N/ 1 o x a l i c a c i d s: O x a lic a cid c r y st a ls c o r r e s po n d to t h e f o r m u l a , ( COOH ) 2 2 H 2 O M o l a r m a ss o f o x a lic a cid c r y st a ls = ( 2  12 + 4  1 6 + 2  1 ) + 2 ( 2  + 1 + 16) =( 2 4 + 6 4 + 2 +3 6 ) = 1 2 6 g m o l - 1 Ba sicity o f o x a lic a cid = 2 So , 126 = 6 3 g e q u iv - 1 E q u i v a l en t m a ss o f o x a lic a cid = 2 T h en, 1 L o f 1 N s o l u t i o n o f o x a l ic a cid c on t a i n s 6 3 g o x a lic a cid 10 m L o f 1 N “ “ “ “ 6 3 g o x a lic a cid N 63 1 × 25 × g o x a li c 25 m L o f ” “ “ “ = a c i d 10 1000 10 63 = = 1 . 57 5 g o x a li c a c i d 1000 18 . Fi n d t h e m o l a r ity a n d m o l a lity o f a 1 5 % s o l u ti o n o f H 2 S O 4 ( d e n s i ty o f H 2 S O 4 s o l u ti o n = 1 . 1 g / c m 3 M o l a r m a ss o f H 2 S O 4 9 8 g m o l -1 ) S olu t ion: 1 5 % S o l u ti o n o f H 2 S O 4 m e a n s t h a t 10 g o f s o l u ti o n s h ou ld c on t a in 1 5 g o f H 2 S O 4 S o . M a ss o f H 2 S O 4 d iss o l v ed = 1 5 g BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 10

11. ADD I T IO N A L PR OB L E M S ON S O L U T I O N S M a ss o f s o l u ti on = 10 g = 1 . 1 g c m - 3 = 9 8 g m o l - 1 D en sity o f t h e s o l u t i on M o l a r m a ss o f t h e H 2 S O 4 So , M a ss o f w a t e r c o n t a i n i n g 1 5 g o f H 2 S O 4 = ( 1 – 1 5 ) g = 8 5 g 1 5 g 15 N o . o f m o l e s o f H 2 S O i n 1 5 g o f H 2 S O = = m o l = .15 3 m ol 9 8 g / m o l 98 m a ss = D en s i t y 100g 90. 9 3 3 and , Vo l u m e o f 1 g o f s o l u ti on = = 90. 9 cm dm 1000 -3 1 . 1 g cm ( i) C a l c ul a t ion of mo l a ri t y N o . o f m o l e s o f H 2 S O 4 0.15 3 m ol -3 M o l a r ity o f H 2 S O 4 s o l u t i o n = = = 1.6 8 m o l dm 3 3 Vo l u m e o f s o l ut i o n i n d m ( 90. 9 / 1000 ) b m So , t h e e m o l a r ity o f H 2 S O 4 s o l u ti o n is 1 . 68 ( ii) C a l c ul a t ion of mo l a li t y N o . o f m o l e s o f H S O 4 2 M o l a r ity o f H 2 S O 4 s o l u t i o n =  1000 M a ss o f w ate r i n g 0.153 -1 -1 M o l a lity = = × 1000 m o l k g = 1 . 8 m o l k g 85 19 . U r e a f o r m s a n i d ea l s o l u ti o n in w a t e r D e t er m i n e t h e v ap o u r p r e s s u r e o f a n a q ue o us s o l u ti o n c on t a i n i n g 1 pe r c e n t b y m a ss u re a a t 4 o C ( Va p ou r p r e s s u r e o f w a t e r o f 4 o C = 55 .3 H g ) S olu t ion: C on c e n t r a ti o n o f u r ea = 10% Let M a ss o f t h e s o l u ti on = 100g T h e n M a ss o f u r ea = 1 g T h e re f o re, M a ss o f w a t er =( 100 - 10 ) g = 90g = 60 g m o l - 1 M o l a r m a ss o f u r e a ( N H 2 CONH 2 ) BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 11

12. ADD I T IO N A L PR OB L E MS n u r ea ON S O L U T I O N S 10/60 1.1667 X = = = u r ea n u r e a + n ( 10/6 + 9 0/1 8 0.166 7 + 5 T h en, M o l e - f r a c ti o n o f u r e a, w ater 0.1667 = 5.1 6 67 X u r ea = 0.032 p o o - p T h en , f r o m R a o u lt’s l a w , = X u r ea p o 55. 3 - P s = 0.032 55. 3 So , t h e v ap o u r o f u r e a s o l ut i o n a t 40 o C is 53 . 5 m m H g . 20 . T h e v ap o u r p r e ss ur e o f w a t e r a t a c re a ti o n t e m p e r a t u r e is 1 8 . 1 5 t o r r . a n d t ha t o f a s o l u ti o n c on t a i n i n g 9 . 4 7 g o f s u g a r , a t t h e s a m e t e m pe ra t u r e , is 1 8 . 6 t o r r C a lc u l a te t h e m o l a r m a ss o f t h e s u g a r . P o S olu t ion Gi v e n Vap o u r p r e s s u r e o f w a t e r ( s o l v en t) = 18 . 1 5 t o r r A Vap o u r p r e s s u r e o f s o l ut i on, P A = 18 . 6 t o r r W B = 9 . 4 7 g W A = 1 0g M a ss o f s u g a r ( s o l u t e) M o l a r o f w a t e r ( s o l v en t) M o l a r m a ss o f w a t e r ( s o l v en t) M A = 1 8 g / m ol M o l a r m a ss o f s u g a r( s o l u t e ) , M B = ? F r o m E q ( 3. 1 1 ), 9.4 7 × 1 8   W B × M A P A 18.03 M B = = g / m o l = 3 4 2 g / m ol o W 10 1 8 .1 5 - 18.06 P - P A A A  o P W × M 9.4 7 × 18  B A A  18.03 F o r m E q . ( 3 . 13 ) M = =  o P   18.1 5 - 1 8 .0 6  W 100 - P B A A A BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 12

13. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 2 1 T h e v apou r p r e ss u r e o f a 5 % a q u e ou s s o l u ti o n o f a n o n – v o l a tice o rg an ic s u b st a n ce a t 3 7 3 K is 7 4 5 m m H g . C a lc u l a t e t h e m o l a r m a ss o f t h e s ub s t an c e . S olu t ion. F r o m t h e g i v e n da t a , f o r 10 g s o l u ti o n, M a ss o f S o l u t e, W 2 = 5 g W 1 =(1 00 - 5 ) g = 9 5 g=0 . 09 5 kg M a ss o f s o l v en t ( w a t e r ) Vap o u r p r e s s u r e o f s o l ut i on , P s = 7 4 5 m m Hg Vap o u r p r e s s u r e o f pu r e s o l ut i on . ( w a t e r ) a t 37 3 K , P o = 76 m m Hg s M o l a r m a ss o f s o l u ti on, M 2 =? M o l a r m a ss o f s o l v en t ( w a t e r), M 1 = 1 8 g / m ol o P - P s n 2 W 2 / M 2 s F o r m R a o u lt,s l a w, = = o P  W  + W n + n / M / M s 1 2 1 1 2 2  5 / M 2  76 - 745 5/ M 2 = = 5.27 8 +  5/ M  o  9 5 /1 8  + 5/ M P 760 s 2 2 T h e a bo v e e q u a ti o n m a y b e r e w ri tt e n as 76 5.27 8 +  5 / M 2  5.27 8 M 2 = = + 1 = 1.0556 M 2 + 1  5 / M  15 5 2 50 . 6 7 = 1 . 05 5 6 M 2 + 1 -1 50.6 7 - 1 o r M = 1.0556 = 47 g m ol 1.0556 2 22 . C a l c u l a te t h e bo ili n g po i n t o f a on e m o l a r a q ueou s s o l u ti o n ( d en sit r y : 1 . 3 g m L -1 ) of s od i u m c h l o r i d e . K b f o r w a t e r = . 5 2 K kg m o r -1 A t o m ic m a s s . Na = 2 3. CI= 35 . 5 . BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 13

14. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S = 1 m o l a r = 1 m o l L - 1 = 1 . 3 · g m L - 1 = ( 2 3 + 35 . 5 ) g / m o l = 5 8 .5 g / m o l - 1 S olu t ion. C on c. o f t h e s o l u ti on D en sity o f s o l u ti o n M o l a r m a ss o f N a Cl So , M a ss o f 1 lit r e o f s o l u t i o n = 100 x 1 . 3 =1 03 g T h e r f o r e, M a ss o f w a t e r c o n t a i n i n g 1 m o le o f N a CI =(1 03 - 58 . 5 ) g = 971 .5 g 1 × 1000 T h u s, M o l a lit y , o f t h e s o l u ti on , m = m o l / k g = 1.029 3 m o l / k g 971. 5 T h en, Δ T b = I  K b  m = (2  . 5 2  1 . 2 9 3 ) K = 1 . 7 K Bo ili n g po i n t o f s o l ut io n = ( 373 . 1 5 +1 . 07 ) K = 37 4 .2 2 K So , 23 . A n a qu eou s s o l u ti o n c o n t a i n i n g 2 . 4 g o f a s ub st a n ce pe r 1 g s o l v en t s ho w s an e l e v a ti o n o f bo ili n g p o i n t o f . 2 1 K w ha t is t h e m o l a r m a ss o f t h e s u b st an c e ? K b f o r w a t e r ( s o l v en t) is . 5 2 k kg m o l - 1 w 2 = 2 .4 g w 1 = 100 g = .1 kg Δ T b = . 2 1 K S olu t ion: M a ss o f s o l u t e. M a ss o f s o l v en t, M o l a r m a ss o f t h e s o l u t e = ? (M 2 ) K b × w 2 0.5 2 × 2. 4 w e k no w , M 2 = = g/ m o l = 59. 4 g / m ol w 1 ×  T b 0. 1 × 0.21 24 . A s o l u ti o n c on t a i n s 3 .5 g o f a n o n - v o l a tile s o l u t e in 12 5 g o f w a t e r ; a n d it bo ils at 373 . 5 2 K . C a lc u l a te t h e m o l a r m a ss o f t h e s o l u t e . ( K b f o r w a t e r = . 5 2 K / m ) S olu t ion · M a ss· o f t h e s o l u t e, M a ss o f w a t e r , E l e v a ti o n o f bo ili n g po i n t, w 2 = 3 :5 g w 1 = 12 5 g = . 12 5 kg Δ T b = ( 37 3 . 52 - 37 3 . ) K = . 5 2 K . = M ( ? ) Le t, M o l a r m a ss o f t h e s o l u te W e k no w t ha t, BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 14

15. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S K  w / M  0.5 2 k g m o l - 1 × 3.5g b K b n 2 2  T b = = w 1 w 1 M × 0.12 5 k g -1 0.5 2 × 3. 5 K k g m o l g . 5 2 K = M × 0.12 5 k g -1 0.5 2 × 3. 5 K k g m o l g 3. 5 -1 g m o l = 2 8 g m ol -1 M = = 0.5 2 K × 0.1 2 5 k g 0.125 25 . T h e b o ili n g po i n t o f w a t e r be c o m e s 10 . 5 2 °C, i f 1 . 5 g o f n o n - v o l a tile s o l u te is d iss o l v e d in 1 m L o f it. C a lc u l a te t h e m o l a r m a ss o f t h e s o l u t e . K b f o r w a t e r = .6 k / m . S olu t ion . M a ss o f s o l u te = 1 .5 g Vo l u m e o f w a t e r ( s o l v en t) = 10 m L T a ki n g d en sity o f w a t e r a s 1 g / m L, M a ss o f w a t e r ( s o l v en t ) = 10 m L x 1 g / m L = 10 g = . 1 k g· Δ T b = (1 00 . 5 2 - 1 00 ) °C = . 52 °C K b w 2 W e k no w , Δ T b = × w 1 K b w 2 M . 6 × 1 .5 or M= × g / m o l = 17 . 3 g / m ol  T b w 1 0.5 2 × 0. 1 26 . A s o l u t i o n o f 3 . 7 9 5 g s u l phur in 1 g c a r bo n d is u l ph i d e (b o ili n g po i n t, 4 6 . 30 ° C ) bo ils a t 46 . 66 °C w ha t i s t h e f o r m u la o f s u l p h u r m o l e c u le in t h e s o l u ti on ? K b f o r c a r b o n d is u l ph i d e is 2 . 4 2 K kg m o l - 1 S olu t ion . M a ss o f s u l p hu r ( s o l u t e ) = 3 . 7 9 5 g M a ss o f c a r bo n d is u l p h i d e ( s o l v en t) = 1 g = .1 kg M o l a r m a ss o f s u l p h u r = M Δ T b = (4 6 . 6 6 - 4 6 . 30 ) ° C = . 36 °C K b w 2 2.4 2 × 3.795 W e k no w, M = = g/ m o l = 2 5 5. 1 g/ m o l w 1  T b 0. 1 × 0.36 BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 15

16. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S A t o m ic m a ss o f s u lp h u r = 3 2 g / m ol So , 255. 1 N o . o f s u l p h u r a t o m s i n o n e m o l e c u le = = 7.9 7 = 8 32 So , s u l phu r e x ists a s S 8 in t h e s o l u t i on. 27 . A s o l u t i o n . o f u r e a i n w a t e r f r ee z e s a t . 40 °C. W h a t w ill b e t h e b o i li n g o f t h e s a m e s o l u ti o n i f t h e d ep r e ss i o n a n d e l e v a ti o n c on s t an ts f o r w a t e r a r e 1 . 8 6 de g kg m o l - 1 a nd .5 1 2 de g kg m o l - 1 r e s pe cti v e l y ? S olu t ion: W e ha v e t h e r e l a t i on s h i p s, K b . n 2 K f . n 2  Tb …. ( i) and  Tf … ( ii) w 1 w 1 w he r e , n 2 is t h e n u m b e r o f m o l e s o f t h e s o l u t e , a n d w 1 is t h e m a ss o f s o l v en t in k g . Divi d i n g E q . ( i) b y E q . ( ii) K  T b b =  T K f f  T f × K b . 4 × . 512 o o C = . 1 1 C or  T b = = K f 1 . 86 Bo ili n g po i n t o f t h e u re a s o l u t i o n = 1 00 o C + . 11 o C = 1 00 . 11 o C T h en, 2 8 . T h e n o rm a l f r ee z i n g po i n t o f n it ro ben z en e , C 6 H 5 NO 2 is 2 4 8 . 8 2 K .A . 2 5 m o l al s o l u ti o n o f c e r t a in i n n i t r oben z en e c au s e s a f r ee z i n g po i n t d ep r e ssi o n o f 2 d e gr ee. C a lc u l a te t h e v a l u e o f K f f o r n it r o b en z ene. S olu t ion: N o r m a l f re e z i n g po i n t o f n it ro ben z e n e C on c e n t r a ti o n o f s o l ut i on = 278 . 8 2 K = . 2 5 m o l a l =0 . 2 5 m o l / kg F r ee z i n g po i n t d e p r e s s i o n  T f = 2K K f  ?  T f = K f ×m W e k no w t ha t, BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 16

17. ADD I T IO N A L PR OB L E M S ON S O L U T I O N S -1  T f 2K K f = = 8 K k g m o l m 0.2 5 m o l / k g 29 . Fi n d t h e ( i) bo ili n g po i n t, a n d ( ii) f re e z i n g po i n t o f a s o l u ti o n c on t a i n i n g . 5 2 g g l u c o se ( C 6 H 1 2 O 6 ) d iss o l v e d in 80 . 2 g o f w a t e r f o r w a t e r . K f =1 . 8 6 K / m , a n d K b =0 . 5 2 K / m . N o r m a l bo ili n g po i n t o f w a t e r = 100 o C N o r m a l f re e z i n g po i n t o f w a t e r = . o C S olu t ion: M a ss o f g l u c o se = . 5 2 g M o l a r m a ss o f g l u c o s e (C 6 H 12 O 6 = 18 m ol 80. 2 M a ss o f w a t e r = 8 .2 g = kg 1000 K n b 2 W e k no w  T = b w he r e , n is n o . o f m o l e s o f s o l u t e ( w / 2 2 w M 2 ) w 1 is t h e m a ss o f s o l v en t in kg 1 0.5 2 × 0.52 0.5 2 × ( w 2 / M 2 )  T b = K = 0.019 K = 0.02K = w 1 18 × ( 80.2/1000) So , Bo ili n g po i n t o f s o l u ti o n = ( 373 + . 2 ) K = . 06 7 K K ( w / M ) 1 . 8 6 × . 52 2 2 f S i m il a r ly  T = f = . 06 7 K 18 × ( 80 . 2 / 1000) w 1 So , F r ee z i n g po i n t o f s o l u t i o n = ( 2 7 3 K – . 6 7 K ) 30 . E t h y l en e g l y c o l ( H o H 2 C- CH 2 OH) is u s e d a s a n a n t i f r ee z e f o r w a t e r to b e u s e d in c a r r ad i a t o r s in c o ld p l a c e s .H o w m u c h e t h y l en e g l y c o l s hou ld b e a dd e d to 1 kg o f w a t er to p r e v en t it f ro m f re e z i n g a t - 10 o ? M o l a l d ep r e ssi o n c on s t a n t o f w a t e r is 1 . 8 6 k kg m o l - 1 S olu t ion: M a ss o f w a t e r ( s o l v e n t) w 1 = 1 kg o  T f = 1 C K f = 1.8 6 K k g m ol w 2 =? K f ( w 2 / 62) -1 M a ss o f e t h y l en e g l y c o l r e q u i r ed, W e k no w ,  T f = w 1 BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 17

18. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 1.8 6 × w 2 1 = 6 2 ×1 1 × 62 × 1 T h is g i v e s w = g = 333.3g 1.86 2 3 1 A s o l u ti o n c o n t a i n i n g 4 g o f a non - v o l a tile o rg an ic s o l u te pe r 10 c m 3 w a s f o u n d t o ha v e a n o s m o tic p r e s s u r e e q ua l t o 5 c m H g . a t 27 o C. C a lc u l a t e t h e m o l a r m a ss o f t h e s o l u t e . S olu t ion: Os m o tic p r e ss u r e T e m p e r a t u re , M a ss o f s o l u t e, Vo l u m e o f s o l u ti on,  = ( 5000 / 7 6 ) a t m T = ( 2 7 + 273 ) =3 K W = 4 g V = 10 c m 3 a tm de g - 1 m o l - 1 R = . 8 2 1 d m 3 a tm d eg - 1 m o l - 1 w W e k no w , p  v = R T M w 4 × 0. 8 21 × 300  50 / 7 6  × 0. 1 T h is g i v e s, M = × R T g/ m ol = 149.7g/ m ol  V 32 . 5 g o f a n o n – v o l a tile non - e l e ct r o l y te s o l u te is d iss o l v e d in w a t e r an d t h e s o l u ti on w a s m ad e u p to 25 c m 3 . T h e s o l u ti o n e x e r t e d a n o s m o tic p r e ss u r e e q ua l t o 4 X 10 5 N M - 2 a t 2 9 8 K Fi n d t h e m o l a r m a ss o f t h e s o l u te S olu t ion: M a ss o f s o l u t e, Vo l u m e o f s o l u ti on, Os m o tic p r e ss u r e T e m p e r a t u re, M o l a r m a ss o f s o l u te w = 5 g = . 5 kg V = 25 c m 3 = 25  1 - 6 m 3 =2 .5  10 - 4 m 3  = 4  10 5 N m - 2 T = 2 9 8 K M =? R= 8 . 31 4 j k - 1 m o l - 1 -1 -1 w R T 0.00 5 k g × 8.31 4 JK m , o l × 2 9 8 K M = = W e k no w , 5 -2 -4 3  V 4 × 1 N m × 2 . 5 × 1 m M= . 123 9 kg m o l - 1 = 1 23 .9 g m o l - 1 or BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 18

19. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 33 . O s m o tic p r e s s u r e o f a s o l ut i o n c o n t a i n i n g 7 .0 g p ro t e in i n 1 m L o f s o l u t i o n is 20 m m Hg a t 3 7 o C C a lc u l a te t h e m o l e c u l a r m a s s o f t h e p r o t e in . ( R = . 08 2 1 lit r e a t m o s p h e r e d e g - 1 m o l - 1 ) S olu t ion: M a ss o f p r o t e i n , Vo l u m e o f t h e s o l u ti o n , w = 7 .0 g V = 10 m L = . 1 L M o l e c u l a r m a ss o f t h e p r o t e in = M T e m p e r a t u re , T = 37 o = ( 37 + 273 ) k =3 1 K 20 Os m o tic p r e ss u re ,  = 2 m m Hg w e ha v e, a tm 760 w R T 7 . × . 082 1 × 310 7 . × . 082 1 × 31 × 760 M = = = = 6770 g / m ol p V ( 20 / 760 ) × . 1 2 × . 1 34 . A t 2 9 8 K , 1 c m 2 o f a s o l u ti o n c o n t a i n i n g 3 . 2 g o f a n u n i d e n t i f i e d s o l u te e x h i b its a n o s m o tic p re ss u r e o f 2 . 5 5 a t m o s p h e r e . W h a t is m o l e c u l a r m a ss o f s o l u t e ? ( R = . 082 1 L l a t m m o l - 1 K -1 ) S olu t ion: Gi v en : T = 2 9 8 K V = 10 c m 3 = . 1 L W = 3 . 2 g  = 2 . 5 5 a t m Vo l u m e o f t h e s o l u ti on M a ss o f S o l u t e, Os m o tic p r e ss u re , w T h e o s m o tic p r e ss u r e is g i v e n b y ,  V = n RT = RT M w R T 3.02 × 0.082 1 × 2 9 8 So . M = = = 6770 g / m ol p V 2 .5 5 × 0. 1 35 . A 5 % s o l u ti o n o f c a n e s u g a r C 1 2 H 2 2 O 1 1 is is o t on ic w ith . 8 7 7 % o f s o l u te A C a lc u l a te t h e r e l a ti v e m o l a r m a ss o f A . A s s u m e den sity o f s o l u ti o n s to b e 1 g c m - 3 S olu t ion: C o n c. o f s u g a r s o l u ti o n = 5% M a ss o f s o l u ti on = 100g So , M a ss o f s u g a r in s o l u t i o n = 5 g R e l a ti v e m o l a r o f s u ga r = ( 1 2  12 +2 2 + 11  16 ) = 144 + 22 + 1 7 6 = 342 W e k no w , nv = n R T BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 19

20. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S n w RT o r,  = R T = × V M V 5 RT  S uga r = So , × 342 V 0.87 7 R T and,  A = × M V S i n c e , t h e t w o s o l u ti o n s a r e is o t on ic, hen c e,  S uga r =  A , So , 0.877 R T 5 R T × = × M V 342 V 0.87 7 × 342 o r, M = = 60g/ m ol 5 36 . C a l c u l a te t h e m o l e c u l a r m a ss o f a s ub st a n ce 1 .0 g o f w h ich o n be i n g d iss o l v e d in 10 g o f s o l v en t g a v e a n e l e v a ti o n o f . 30 7 K in t h e b o ili n g po i n t. ( M o l a l e l e v a ti on c on st a n t, K b = 1 . 8 4 K / m ) . K b × n 2 S olu t ion: W e k no w ,  T b = w he r e n 2 is t h e no . o f m o l e s o f s o l u t e ; w 1 is w 1 t h e m a ss o f t h e s o l v en t in k g. M a ss o f s o l ut e 1. g n 2 = = M o l a r m a ss o f th e s o l ut e M and w 1 = 10 g = 0. 1 k g Sub sti t u ti n g t h e se v a l u e s, o n e c a n w ri te, 1.84 K k g m o l -1 1 . g 0.307K = × 0.1K g M 1.84 K k g m o l - 1 × 1. 1.8 4 × 1. g So , M = × g / m o l= 5 9 .9 g / m ol 0.1K g × 0.3 7 K 0. 1 × .307 37 . T h e d en sity o f w a t e r is r oo m t e m p e r atu r e i s 09 9 7 g c m 3 C a lc u l a t e t h e m o l a r ity o f pu r e w a t e r . S olu t ion: T h e d en sity o f w a t e r is . 99 7 g c m 3 . T hu s, BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 20

21. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S M a ss o f 1 c m 3 o f w a t e r = . 9 9 7 g T h e re f o re , M a ss o f 1 d m 3 =1 00 c m 3 o f w a t e r = . 9 9 9 7 g c m - 3  10 c m 3 = 9 9 7 g M o l a r m a ss o f w a t e r = 2 + 16 = 1 8 g m o l - 1 N o . o f m o l e s o f w a t e r pe r d m 3  55.39 T h e re f o re, T h e re f o re , t h e m o l a r ity o f w ate r is 5 5 . 3 9 m o l L - 1 38 . A s o l u t i o n is 2 5 pe r c en t w a t e r 2 5 p e r c en t e t h a no l a n d 5 pe r c e n t a c e tic a cid by m a ss. C a lc u l a t e t h e m o l e - f r a cti o n o f ea c h c o m p o n e n t. S olu t ion: L e t u s c o n si de r 1 g o f t h e s o l u t i o n . T h e n, M a ss o f w a t e r , M a ss o f e t h a no l, M a ss o f a c e tic a c i d m w = 2 5 g m e t h = 25g m a c e t = 5 g M o l a r m a ss o f w a t e r , e tha n o l a n d a c i d a re , 1 8 g / m o l, 4 6 g / m o l, a n d 6 g / m o l r e s pe cti v e ly T he n , 25g N o . o f m o l e s o f w a t e r , n w = = 1.39 1 8 g/ m ol 25g N o . o f m o l e s o f e t ha no l, n e t h = = 0.54 46g/ m ol 50g N o . o f m o l e s o f a c e tic a cid n a c e t = = 0.83 60g/ m ol =( 1 . 3 9 + . 54 + . 83 ) =2 .76 1.39 T o t a l n o o f m o l e s in s ol u ti on So , M o le – f r a cti o n o f w a t e r , X w = = 0.50 2.76 0.54 M o l e - f r a c ti o n o f e t h a n o l, X e t h = = 0.20 2.76 0.83 M o l e - f r a c ti o n o f a c e tic a cid X a c e t = = 0.30 2.76 39 . C o n c e n t r a t e d s u l p hu r ic a cid ha s a d en si t y o f 1 .9 g / M L an d is 9 9 % H 2 S O 4 b y w e i g h t C a lc u l a te t h e m o l a r ity o f H 2 S O 4 in t h is a ci d. S olu t ion: Le t u s c o n s i de r 1 lit r e o f H 2 S O 4 T h en M a ss o f 1 lit r e o f t h is a c i d = 100 m L  1 .9 g / m L = 19 g S i n c e , t h e g i v e n a cid is 99 % pu r e , h e n ce BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 21

22. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 190  99 A ct ua l m a ss o f H 2 S O 4 in 1 lit r e s a m p l e = = 188 1 g 100 M o l a r m a ss o f H 2 S O 4 = ( 2 + 32 + 64 ) g m o l -1 = 9 8 g m o l - 1 188 1 g So , No o f m o l e s o f H 2 S O 4 in 1 lit r e o f t h e s a m p le = = 19.1 9 m ol -1 9 8 g m o l So , t h e g i v e n H 2 S O 4 s a m p l e is 1 9 .9 m o l a r . 40 . C o n c e n t r a t e d n it r ic a c i d u s e d a s a l abo r a t o r y r ea g en t is u s ua lly 69 % b y m a ss o f n it r ic a cid C a lc u l a te t h e v o l u m e o f t h e s o l u ti o n w h ich c on t a i ne d 2 3 g o f H NO 3 . D en sity o f t h e c on c . H NO 3 So l u ti o n is 1 . 4 1 g c m -3 . S olu t ion: Le t, M a ss o f c on c. HNO 3 s a m p le =1 00 g So , and M a ss o f HNO 3 in 1 g o f s a m p l e = 6 9 g M a ss o f HNO 3 in 1 g o f s a m p l e = 3 1 g D en sity o f c on c . HNO 3 s a m p le = 1 . 4 1 g c m - 3 M a ss 2 100g So , V o l u m e o f 1 g o f th e H NO 3 s a m p le = = = 70.92 cm c m -3 D w n s i t y 1.41g T h u s, 6 9 g o f HNO 3 is c on t a i ne d i n . 7 . 9 2 c m 3 o f c o n c. HNO 3 70.92 1g “ “ “ 69 70.92 3 23g “ “  2 3 “ = 23 .6 c m 69 T h u s, 2 3 .6 c m 3 c o n c. HNO 3 s a m p l e c o n t a i ne d 2 3 g HNO 3 41 . A So l u ti o n c on t a i n i n g 4 .2 gr a m s o f a n o rg an ic t h e m o l a r m a ss o f t h e o rg an ic c o m p o u n d i n K b o f a c e t o n e = 1 . 7 1 k kg m o l - 1 S olu t ion: M a ss = o rg an ic c o m p ou n d, M a ss o f a c e t o ne , ( S o l v en t ) . W 2 = 4 .2 g W 1 = 50g Δ T b = 1 . 8 K b = 1 . 7 1 k kg m o l - 1 If t h e m o l a r m a ss o f t h e s o l u t e is M , t hen BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 22

23. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 100 × K b × w 2 100 × 1.7 1 × 4. 2 -1 -1 M = = g m o l = 79. 8 g m ol w 1 × V T b 50 . × 1. 8 42 . C a lc u l a te t h e m o l al ity o f a K CI s o l u ti o n i n w a t e r s u ch t h a t t h e f r e e z i n g po i n t is dep r e s s e d b y 2 K . ( K f f o r w a t e r = 1 . 8 6 K kg m o l -1 ) S olu t ion: F o r K C I , t h e f r ee z i n g po i n t d ep r e s s i o n is g i v e n b y ,  T f = i K f m  T f 2 -1 -1 T h is g i v e s, m = = m o l k g = 0.54 m o l k g i K f 2 × 1.86 4 3 . A de ci n o r m a l s o l u ti o n o f N a CI e x e r ts a n o s m o tic p r e s s u r e o f 4 . 6 a t m a t 3 K . c a lc u l a te its de gr e e o f d i ss o ci a ti o n . ( R = . 08 2 L a tm K - 1 m o l -1 ) S olu t ion: N o . o f m o l e s o f N a CI p e r lit r e o f s o l u ti o n = .1 Os m o tic p r e ss u re ,  = 4 .6 a t m T e m p e r a t u re, T = 3 K H a d N a CI n o t d iss o ci a t e d t hen  no r m a l = CR T = 0. 1 × 0.08 2 × 300at m = 2.46at m Bu t  ob s = 4 . 6 a tm A s pe r d iss o ci a ti o n O b s e r v ede m agn i tudeofa c o lli gatu v ep r ope r t y p ob s = 4.6at m i = = = 1.87 p nor m al N o r m a l m agn i tudeofa c o lli gat i v ep r ope r t y 2.46at m F o r t h e t h e d iss o ci a t i o n o f a n e l e ct r o l y te p ro du ci n g n io n s, i  1 1.8 7  1  1 0.87 a   0.87 n  1 2  1 1 So , P r c e n t a g e d iss o ci a t i o n = 10   = 10  . 8 7 =8 7 % 44 . T h e d e gr e e o f d is s o ci a ti o n o f Ca ( NO 3 ) 2 in a d il u t e a q u e ou s s o l u t i o n c o n t a i n i n g 7 .0 g o f t h e p e r 1 g o f w a t e r a t 1 o C 7 0pe r c en t. If t h e v apou r p r e ss u r e o f w a t e r a t 100 o C is 76 m m , t h e m c a l c u l a te t h e v apou r p r e s s u r e o f t h e s o l u t i on’ S olu t ion: E a ch m o l e c u le o f c a lci u m n it r a t e , ( C a ( NO 3 ) 2 g i v e n 3 i on s o n d iss o ci a ti o n in s o l u ti o n e . g ., BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 23

24. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S C a (N O ) 2  C a 2 + + 2 NO - 3 3 So , n = 3 . F o r e l e ct r o l y t e s w h ic h - d iss o ci a te i n to i on s in So l u t i on, i -1  = n -1 70 i =  ( ν - 1 ) + 1 = o r, × ( 3 - 1 ) + 1 = 2. 4 100 N o r m a l m o l a r m a ss o f C a ( N 3 ) 2 = ( 4 + 2 x 1 4 + 6 x 16 ) g / m o l = 16 4 g / m o l F r o m t h e R a ou lt's l a w o p -p n 2 = i o n +n 1 2 p F o r d il u te s o l u ti o n , n 2 < < n 1 . So .    i  w /M  76 - P  2  2. 4 × ( 7/164) 2. 4 × 7 × 18 2 = = = = 0.0 1 84 760 w / M 1 1 ( 1000/1 8 ) 1 6 4 × 100 760 - p = 7 6 x o . 1 8 4 m m Hg = 1 4 m m Hg p = ( 760 -1 4 ) m m Hg = 74 6 m m Hg 45 . C a l c u l a te t h e a m o un t o f K Cl w h ich m u st b e a dd e d t o 1 kg o f w a t e r so t hat t h e f r ee z i n g po i n t is d ep r e ss e d b y 2 K . ( K f (w a t e r ) = 1 . 8 6 K kg m o l - l ) S olu t ion: M a ss o f w a t e r , w 1 = 1 kg M a ss o f K Cl P e r kg o f w a t e r .= w 2 g M o l a r m a ss o f K CI = ( 3 9 + 3 5 . 5 ) g m o l - 1 = 7 4 .5 g m o l - l  i K ×  w / 74 5  n   f 2 2 W e k no w , ΔT f = I K f  m = i K f  = w 1 1 F o r K CI, i =2  T × 745 f 2 × 74 . 5 74 . 5 g = w = = 40 . 05g i K 2 × 1 . 8 6 1 . 86 2 f BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 24

25. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S 46 . C a l c u l a te t h e a m o un t o f s o d i u m c h l o r i d e w h ich m u st b e a d d e d t o 1 m L o f w a t er so t h a t its f re e z i n g po i n t is d ep r e ss e d b y . 7 4 4 K F o r w a t e r K f =1 . 8 6 K / M A ss u m e den sity o f w a t e r to b e on e g m L - 1 S olu t ion: M a ss o f N a CI , w 2 =? Vo l u m e o f w a t e r = 10 m L ΔT f = . 7 4 4 K K f ( W a t e r ) = 1 . 8 6 K /m D en sity o f w a t e r = 1 g m L - 1 M a ss o f w a t e r , w 1 = 100 m L  1 g m L -1 =1 00 g = 1 kg So ,  w 2 / 58 . 5   × = i K f w 2 A cc o r d i n g to t h e d e f i n i t i on ,  T = i K m = i k       w 1 58. 5 × w 1 f f f  T × 58. 5 × 1 f 0.74 4 × 58. 5 w = = = 11.7g i K 2 × 1.86 2 f So M a ss o f N a CI r e q u i r e d = 11 . 7 g 47. C a lc u l a te t h e a m o u n t o f s o d i u m c h l o r i d e ( e l e ct r o l y t e ) w h ich m u st b e ad d e d to on e kil o gr a m o f w a t e r so t h a t t h e f re e z i n g po i n t is d ep r e ss e d b y K G i v e n : K f f o r w a t e r = 1 . 8 6 K kg m o l - 1 S olu t ion: M a ss o f s o d i u m c h l o r i de . w 2 =? M o l a r m a ss o f s od i u m c h l o ri de. M a ss o f w a t er M 2 = 58 .5 g / m ol w 1 = 1 kg K f = 1 . 8 6 K kg m o l - 1 I = 2 F o r N a CI  w 2 × M 2  2 × 1 . 8 6 × w 2 So ,  T f = i K m = i K × = w 1 58 . 5 × 1 f f  T f × 58 . 5 3 × 58 . 5 o r, w = = g = 47 . 18g 2  1 . 86 2 × 1 . 86 2 48 . T h e f re e z i n g po i n t d ep r e ssi o n . o f o . 1 m N a Cl s o l u ti o n is . 3 72 °C. W ha t c o n cl u si on w ou ld y o u d r a w abou t i ts m o l e c u l a r s t a t e ? K f f o r w a t e r is 1 . 8 6 K kg m o l - 1 S olu t ion: M o l a lity · o f N a Cl s o l u ti on , m = .1 ΔT f , = . 3 7 2 °C K f = 1 . 8 6 K kg m o l - 1 ΔT f = i K 1m . 3 7 2 = i X 1 . 8 6 X .1 . 372 F o r i on ic s ub s t an c e s , w e ha v e or i = = 2 1 . 8 6 × . 1 BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 25

26. ADD I T IO N AL PR OB L E M S ON S O L U T I O N S T h e v a l u e o f i = 2 , i n d ic a t e s t ha t t h e s o l u te N a CI in s o l u ti o n is c om p l e t e ly d iss o ci a t e d g i v i n g t w o i on s, i. e ., N a CI  N a + ( a q ) + Cl - ( a q). 49 . T h e v ap o u r p r e s s u r e o f a p u r e li q u id A is 4 m m Hg a t 3 1 K T h e v apou r p r e ss u r e o f t h is li q u id in a s o l ut i o n w ith li q u id B is 3 2 m m Hg . C a lc u l a t e t h e m o l e f r a cti o n o f A in t h e s o l u ti o n i f it obe y s t h e R a ou lt’s l a w . o S olu t ion: V a p ou r p re ss u r e o f pu r e A , P A = 4 m m Hg Vap o u r p r e s s u r e o f A s o l u ti o n , P A = 3 2 m m Hg o A cc o r d i n g to t h e R aou l t ’ s l a w , P A = P A X A P A 3 2 m m H g T h en, X = =  .8 o P A 4 m m H g A 50 . A s o l u t i o n o f 1 2 .5 g u r e a in 1 7 g u r e a in 17 g o f w a t e r g a v e a bo ili n g po i nt e l e v a ti o n o f . 6 3 K C a l c u l at e t h e m o l a r m a ss o f u re a , t a ki n g K b = . 5 2 K / m . S olu t ion : Gi v en : M a s s o f u re a , o f u r ea , w 2 = 3 .5 g M a ss o f w a t e r , w 1 = 17 g= . 1 7 kg E l e v a ti o n o f bo ili n g po i n t, Δ T b = . 6 3 K M o l a r m a ss o f t h e s o l u te ( u r ea ) is g i v e n b y , K × w . 5 2 × 12 . 5 b 2 g / m o l = 6 .7 m o l - 1 M = = w × V T . 1 7 × . 63 1 b BA NG A L ORE I NS T I T U TE OF C O A C H I N G II P UC C H EMIS T R Y STU D Y M A TE R IAL Pa ge 26