Question

Vapour Pressure of Solutions of Non-Volatile or Volatile Solutes 22.555 g of a non-volatile solute is dissolved in 215.0 g of water. The solute does not react with water nor dissociate in solution. Assume that the resulting solution displays ideal Raoult's law behaviour. At 10°C the vapour pressure of the solution is 9.073 torr. The vapour pressure of pure water at 10°C is 9.209 torr. Calculate the molar mass of the solute (g/mol). See example 17.1 on pp865-6 of Zumdahl ''Chemical Principles'' 7th ed. 1pts Submit Answer Tries 0/5 Now suppose, instead, that 22.555 g of a volatile solute is dissolved in 215.0 g of water. This solute also does not react with water nor dissociate in solution. The pure solute displays, at 10°C, a vapour pressure of 0.921 torr. Again, assume an ideal solution. If, at 10°C the vapour pressure of this solution is also 9.073 torr. Calculate the molar mass of this volatile solute.

Answer 1

vapor pressure of water at 10 deg.c= 9.21 torr

Vapor pressure of compnent in solution = mole fraction of the componemet * pure componet vapor pressure

moles of water =215/18 ( 18 is molar mass of water )=11.94

let M= molat mass of non-volatile solute, its molar mass = M, hence moles = 22.555/M

total moles of mixture = 11.94+22.555/M

Mole fraction of water = 11.94/(11.94+22.555/M)

hene from Raoult's law, 9.209 = 11.94/(11.94+22.555/M)*9.21

hecne 11.94/ (11.94+22.555/M)= 9.209/9.21= 0.999

11.94 = 0.999*(11.94+22.555/M)

11.94= 11.94*0.999 +22.555*0.999/M

0.01194 =22.53/M

M= 22.53/0.01194 =1887 gm/mole

2.   when the solute is also voltatile

equilibruim pressure = mole fraction water* pure component vapor pressure + mole fraction voltaile component* its pure compnent vapor pressure

9.073= 11.94/ (11.94+22.555/M)* 9.21 + 22.555/(11.94+22.555/M)* 0.921

9.073= (11.94*9.21+22.555*0.921)/(11.94+22.555/M)=131/ (11.94+22.555/M)

11.94 +22.555/M= 131/9.073= 14.44

22.555/M= 14.44-11.94= 2.5

M= 22.555/2.5= 9 gm/mole