Find molar mass of nonvolatile solute when 20 g of it dissolved in 200 g water at 300 K . [Relative lowering of vapour pressure = 0.02 ]

Question

TestHub · Accepted Answer

The relative lowering of vapour pressure for a dilute solution is given by the formula:

$\frac{Δ P}{P ^{0}} = \frac{W _{2} M _{1}}{M _{2} W _{1}}$

where:
$W_{2} = 20 g$ (mass of solute)
$W_{1} = 200 g$ (mass of solvent, water)
$M_{1} = 18 g mol^{- 1}$ (molar mass of water)
$\frac{Δ P}{P ^{0}} = 0.02$ (relative lowering of vapour pressure)

Substituting the given values into the formula:

$0.02 = \frac{20 \times 18}{M _{2} \times 200}$

$0.02 = \frac{360}{200 M _{2}}$

$0.02 = \frac{1.8}{M _{2}}$

$M_{2} = \frac{1.8}{0.02} = 90 g mol^{- 1}$

Answer: $90 g mol^{- 1}$

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