LIQUID SOLUTION EXERCISE # 2

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Janice Adelia Lyons
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1 LIQUID OLUION EXERCIE # 4. m d of solute. 4.. () he concentration of solution (ppm) t. of solute 6 t.of solvent ppm 6. m k.69.5 C.69.5 C solvent.59 C C.59 C.789 C. i k f [NaCl Na + + Cl ].86 i.7 solvent.7 O.7.7 C. m k. k k %.8 4. / i + / % % solution means ml solution contain 5 g cane sugar.877 % means ml solution contain.877 g X isotonic solution C C W W Osmotic pressure Colligative properties O icr lcl (i 4), acl (i ), Urea (i ) LIQUID OLUION EXERCIE # K k m.5.5 m.9 K f m' m' m/ K k f g/mole 6. k.7 i i.54. i.5 i + n.5 + n n 9. i k m x y x + + y + i + + i i k m K.7 C 8. cetone gm ater 9 gm / 58 mole % of acetone / 58 9 / 8.% 9. K f m m loss in t.of solvent loss in eight of solution from eq. (i) and (ii) molal... (ii)...(i)

2 . Given that, 64 mm, s 6 mm,.75 g, W 9. g, 78 s m W s m W m Given K ' f m W W t. of enzene V d g m m LIQUID OLUION No.64 g, K f ' 5. K mol m kg for I case...i Weight of solvent ()9+88g m for II case...ii y eq. (i) m.85 m y eq. (ii) m.5 m m.5 m On sustituting in Eq. (i), mm K fw 6. W gm/mol. K f olality ( + ) For acetic acid : CH COOH CH COO +H + Given,. ; lso, molality () EXERCIE # 4[] mole of acetic acid ( ) eight of ater in kg ( kg ) K f olality ( + ) i K f m.6 i.86. i.. 4 i n i K m.46 i.5 i For. solution, i n V n n / V...8 K.46 atm For. solution, V n n / V...8 K.46 atm he movement of solvent particles occurs from dilute to concentrate solution, i.e.,. to. solution. hus, pressure should e applied on concentrated solution, i.e., on. solution to prevent osmosis.,..,. lso, magnitude of external pressure ( ) atm pressure on. solution.

3 6. For initial solution, 5 atm, 8 K 76 5 V n (i) fter dilution, let volume ecomes V and temperature is raised to 5 C, i.e, 98 K (V 5 C 98 K) 5. atm V n (ii) LIQUID OLUION. ik m.5.5 m. No, (NO ) + NaCl Cl + NaNO No, this solution contains to salts () K f.8.86 [. + s] here s is molar soluility of Cl. (s, Cl ) s.54 K sp 4s i K (x ) (x ) (i) x... (ii) From equation (i) and (ii) ( (i) (ii) ) x.4 Formula of protein () H amu. + n + X + Let a mole of are left due to polymerization after min. ( min. a.) a 5 a a. k log a... (i)... (ii) after minute solute is added & final vapour pressure is 4 mm Hg i.e. s 4 ( 4 mm Hg s 4 ) y Eqs. (i) and (ii), e get V 8 5. V V 98 5 V 5 V 5 V i.e., solution as diluted to 5 times. ( 5 ) 8. i c i n.46 i 5 i atm (a ) EXERCIE # 4[]... (iii) from equation (i) and (iii) a 9.9 putting this in eq. (ii) k ((ii) k). log eaker () :-. 4 ole fraction of urea ( ) eaker () : ole fraction of glucose ( ) ole fraction of glucose is less so vapour pressure aove the glucose solution ill e higher than the pressure aove urea solution, so some H O molecules ill transfer from glucose to urea side in order to make the solutions of equal mole fraction to attain equilirium. Let x mole of H O transfered ( H O x H O )

4 x x x 4 no mass of glucose solution ( ) t. % of glucose (%) % 5. Let n mole of present in mole of mixture that has een vaporized. ( n ) hus, Y n 7. C H 5 OH V ml, d.789 g/ml m g H O V 4 ml, d.997 g/ml m g otal mass g d sol..957 g/ml V sol ml % change m % X + X + ( X ) + X ( ) n X + + ( ) Y X ( ) n...(i).4x.4..8x Y X n n n ( n )..8 X n from equation (i) and (ii)...(ii) so atm 6. (a) on solving 9 mm Hg f k f k C () Relative loering of vapour pressure (c) n n N CR n 5.5. v 6.8 ml atm (d).6 W 5.5 W ( + ) % V 75 ml L V 77 ml 89.9 L ( ) X Y torr.7.7

5 .84. i C.,, X Y 4 On condensation X 4, Y 5 5. on further condensation X.. CH OH V ml, d.798 g/ml m.94 g H O V 7 ml, d.9984 g/ml m g m 9.88 g d solution.9575 g/ml V solution 98 ml f 9.9 C , 9 n , n X Y Y.85 On further condensation.85, X Y For to immiscile liquids d ln d H R 4.97.%...(i) log log ln ln d ln d (ii) Compairing equation (i) & (ii) H R[ ] at 8 K H Cal. 7. s.6 s.6 n ~ n n N N moles of solute moles of H O.6 mole H O 8 g 8 ml 8 ml solution.6 mole L solution.6.5 mole/l 8 Let soluility of salt 4 is s then 7s.5 s.5 mole/l k sp. 4 4 (s) (.5) 7 k sp t C : For C 6 H 6 V ml.877 V mm Hg It vapour pressure of enzene at 7 C is then ln H V R ln mmhg m s s solvent m k f C C f

6 9. Initial moles of H O.9 6 kj. k f 8.4 (78.4) k f Rf H f 8.4 (7) m so k f m in g H O.75 mole solute in g H O.75 mole solute in.9 8 g H O mole solute mole of solute (n).74.5 i k f i C C C C/ C / k (C C ) C( ) s n N s k.64(.47).5 moles of H O (N) moles of Ice separate out mass of Ice separate out g. ( + ) k f. ( + ) C k a.9(.58) NH 4 ClNH 4 + +Cl, NH 4 + +H O NH 4 OH+ H + C C C C C CC Ch Ch Ch i C C C + C C h C h C h C ( + + h) i k f.67 ( h) h.8 & since.75, h.9..5 [ + ].6 C...(i).7.86 C ( + ) C + C. from equation (i) & (ii) C.684,.876 C k a.684(.876) (.876)...(ii) 7. In 6 ml solution [F]C C.68 mole/l so moles moles left after 4.8 years.88 4 moles disinitegrated moles of -particle emittted No. of -particle emittted

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www.tutor-homework.com (for tutoring, homework help, or help with online classes) 1. If heat is added to ice and liquid water in a closed container and, after the addition of the heat, ice and liquid water