16.636 g of a non-volatile solute is dissolved in 410.0 g of water. The solute d

Question

16.636 g of a non-volatile solute is dissolved in 410.0 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 90°C the vapour pressure of the solution is 517.01 torr. The vapour pressure of pure water at 90°C is 525.80 torr. Calculate the molar mass of the solute (g/mol).

answer = 43.0 g/mol

Now suppose, instead, that 16.636 g of a volatile solute is dissolved in 410.0 g of water. This solute also does not react with water nor dissociate in solution. The pure solute displays, at 90°C, a vapour pressure of 52.58 torr. Again, assume an ideal solution. If, at 90°C the vapour pressure of this solution is also 517.01 torr. Calculate the molar mass of this volatile solute.

Explanation / Answer

we have formula

Vapor presur of solution = ( mol fc of solute x Vapor pressure of pure solute ) + ( vapor pressure ofs olvent x vapor pressure of pure solvent)

let mol fraction of solute = X , then solvent mol fraction = 1-X

517.01 = ( 52.8 X) + ( 1-X) ( 525.8)

8.79 = 473X

X = 0.0186 = moles of solute / ( moles of solute + water moles)   ,

let moles of solute = m , water moles = 410 /18.015 = 22.76

0.0186 = ( m / m + 22.76)

0.0186m + 0.4233 = m

m = 0.4313 = mass of solute / molar mass of solute

0.4313 = ( 16.636 / Molar mass of solute )

Molar mass of solute = 38.6 g/mol

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