Question

29. The vapour pressure of a glucose solution, C6H1106, is 17.01mmHg at 2000, while the vapour pressure of pure water at 200C is 17.25mmHg. Calculate the molar fraction of the solution. 30. Calculate the molality of an aqueous glycine solution that freezes at -1.10. Kfs 1,86C/m

Need the answers of these two problems please :)

Answer 1

Solutim:-29). It is based on colligative properties Let Ilap our pressure of pure water = P² = 17:25mming and Vapour pressure of glucose soluth = p 17:01 mm Hg ss use we know, - ps Relative lowering of - = - هاه (۳) هیاه % fraction solute) lapour pressure it X Solute = 0.24 O.OY 17.25 -17.01 17:25 sa 17.25 Note :- Xsolute t X Solvent = solvent = 1 -0.014 = 0.986 i molar fraction of solution is the mole component at present present in solution fraction of mole fraction mole fraction of of solute = 0.014 solvent = 0.986 > and

solution 3o) ke = 1.86 Clm , • Aqueous solution of glycine freezes at -11°C le. Ates lol K 0-6 lol) = l'1°C C Difference remain as told • We know, atrai Kfm @ (Nant Hoff factor) i=1 (forglyreine) ma molality of solution. • It is concept of depressing in freezing point Normally , water freezes 0°c but when solute is added it freezes at some below temperature, which is called depression in freezing point. and this is difference is (A TIJ. o now i using eq. Hila 186 x4 xm ma lo! - 0059 lapolal 1086 & solution is molality of aqueous 0.591 molal.
Please give me better rattings.

Thank you.