Question

16.177 g of a non-volatile solute is dissolved in 290.0 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 40°C the vapour pressure of the solution is 54.506 torr. The vapour pressure of pure water at 40°C is 55.324 torr. Calculate the molar mass of the solute (g/mol).

Now suppose, instead, that 16.177 g of a volatile solute is dissolved in 290.0 g of water. This solute also does not react with water nor dissociate in solution. The pure solute displays, at 40°C, a vapour pressure of 5.532 torr. Again, assume an ideal solution. If, at 40°C the vapour pressure of this solution is also 54.506 torr. Calculate the molar mass of this volatile solute.

Answer #1

**raoults law**

** P0-P/P0 = i*Xsolute**

** P0-P/P0 = n2 / n1+n2**

**i = vanthoffs factor of solute = 1
P = vapor pressure of water above the solution = 54.506
torr**

**p0 = vapor pressure of pure water at this temperature =
55.324 torr**

** n1 = no of mol of solvent = (290/18) =
16.11 mol**

** n2 = no of mol of solute particles =
(16.177/x) mol**

**(55.324-54.506)/55.324 =
((16.177/x)/((16.177/x)+16.11)**

**x = molarmass of solute = 66.91 g/mol**

**from daltons law**

**Ptotal = Psolute + pSolvent**

** =
Xsolute*P0Solute + Xsolvent*P0solvent**

** 54.506 =
((16.177/x)/((16.177/x)+16.11)*5.532 +
((16.11)/((16.177/x)+16.11)*55.324**

**x = molarmass of solute = 60.11 g/mol**

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Assume that the resulting solution displays ideal Raoult's law
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At 40°C the vapour pressure of the solution is 54.453 torr.
The vapour pressure of pure water at 40°C is 55.324 torr.
Calculate the molar mass of the solute (g/mol).
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