An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

Solution

Answer:

Given values

Concentration is given in percent so that take

mass of the solution = 100 g

mass of non-volatile solute = 2% = 2g mass of the solvent = (100 — 2) = 98 g molecular mass of solvent (water) = 18

We have to find molecular mass of solute

The vapour pressure of pure boiling water = 1atm = 1.013 bar.

Change in vapour pressure = (1.013 — 1.004)

= 0.009 bar

Formula of Raoult’s law

\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{n_{2}}{n_{1} + n_{2}}

Here n1 and n2 are the number of moles of solvent and solute respectively present in the solution. For dilute solutions n2 < < n1, hence neglecting n2 in the denominator we have

\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{n_{2}}{n_{1}}

Use this formula we get

\frac{p_{1}^{0} - p_{1}}{p_{1}^{0}} = \frac{w_{2} \times M_{1}}{M_{2} \times w_{2}}

Here w₁ and w₂ are the masses and M₁ and M₂ are the molar masses of the solvent and solute respectively.

Plug the values in above formula we get

\frac{0.009}{1.013 bar} = \frac{2 g}{M_{2}} \times \frac{18 g m o l^{- 1}}{98 g}

Cross multiply we get

M_{2} = \frac{2 \times 18}{98} \times \frac{1.013}{0.009} g {mol}^{- 1} = 41.35 g {mol}^{- 1}

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Solutions

An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?

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