A) 14.028 g of a non-volatile solute is dissolved in 360.0 g of water. The solut

Question

A) 14.028 g of a non-volatile solute is dissolved in 360.0 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 70°C the vapour pressure of the solution is 230.70 torr. The vapour pressure of pure water at 70°C is 233.70 torr. Calculate the molar mass of the solute (g/mol).

B) Now suppose, instead, that 14.028 g of a volatile solute is dissolved in 360.0 g of water. This solute also does not react with water nor dissociate in solution. The pure solute displays, at 70°C, a vapour pressure of 23.37 torr. Again, assume an ideal solution. If, at 70°C the vapour pressure of this solution is also 230.70 torr. Calculate the molar mass of this volatile solute.

Explanation / Answer

A)

moles of solute = n1

moles of solvent water = n2 = 360 / 18 = 20

pure solvent vapour pressure = Po = 233.70 torr

solution vapour pressure = Ps = 230.70

relative lowering vapour pressure = Po -Ps / Po

                                     = 233.70 - 230.70 / 233.70

                                    = 0.0128

from raoult's law :

relative lowering vapour pressure = mole fraction of solute

0.0128   = n1 / n1 + n2

0.0128 = n1 / n1 + 20

n1 = 0.259

moles of solute =0.259

mass of solute = 14.028 g

molar mass = 14.028 / 0.259

molar mass = 54.1 g/mol

B)

this problem is given wrong pure solute vapour pressure = 23.37 torr less than that of solution vapour pressure = 230.70 torr . this should not be correct check once

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