CBSE Class 12 · 2005 · Foreign · SolutionsShort Answer

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A solution is made by dissolving 30 g of a non-volatile solute in 90 g of water. It has a vapour pressure of 2.8 kPa at 298 K. At 298 K, the vapour pressure of pure water is 3.64 kPa. Calculate the molar mass of the solute.

Answer: M = 20 g mol⁻¹ (*)

Solution

By Raoult's law for a non-volatile solute,

p = x_{1} p^{\circ}

x_{1} = \frac{p}{p ^{\circ}} = \frac{2.8}{3.64} = 0.7692

Moles of water:

n_{1} = \frac{90}{18} = 5 mol

From

x_{1} = \frac{n _{1}}{n _{1} + n _{2}}

0.7692 = \frac{5}{5 + n _{2}} \Rightarrow 5 + n_{2} = 6.5 \Rightarrow n_{2} = 1.5 mol

Molar mass:

M = \frac{30}{1.5}

M = 20 g mol^{- 1}

(*) The data as printed give this unusually small value; the exact mole-fraction relation has been used throughout.

CBSE Class 12 Board · 2005 · Foreign · Subjective

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