5.661 g of a non-volatile solute is dissolved in 235.0 g of water. The solute do

Question

5.661 g of a non-volatile solute is dissolved in 235.0 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 60°C the vapour pressure of the solution is 148.36 torr. The vapour pressure of pure water at 60°C is 149.40 torr. Calculate the molar mass of the solute (g/mol). ANSWER = 61.9 g/mol

Now suppose, instead, that 5.661 g of a volatile solute is dissolved in 235.0 g of water. This solute also does not react with water nor dissociate in solution. The pure solute displays, at 60°C, a vapour pressure of 14.94 torr. Again, assume an ideal solution. If, at 60°C the vapour pressure of this solution is also 148.36 torr. Calculate the molar mass of this volatile solute. ANSWER = ?????

Explanation / Answer

Assume that

a = water, b = solute.

Pt = xa (Poa) + xb (Pob)

Mole total = Mole fraction xa + xb = 1 so xb = 1 - xa

148.36   torr = xa( 149.40 torr) + (1 - xa)(14.94 )

148.36   = 149.40 xa + 14.94 - 14.94 xa

133.96 = 134.46 xa

xa = 0.996

xb =1 -0.996 = 0.004

moles H2O = 235 g(1mol/18.02g) = 13.04 moles

Total moles = moles H2O / mole fraction H2O = 13.04/ 0.996

= 13.09 moles

Moles solute = Total moles - moles H2O = 13.09 – 13.04 = 0.05 moles

0. 05 mol/5.661 g = 1mol/x grams = 113.22 g/mol

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