Name: Azeotropic Mixtures
Uploaded: 2019-07-19
Duration: 8 min 23 s
Description: Certain liquid mixtures have a definite composition and boil at a constant temperature. They are called azeotropes. Let’s explore the cause of their behaviour!

Question

The vapour pressure of a $5 %$ aqueous solution, by mass, of a non-volatile organic substance at $373 K$ is $745 mm$ of $Hg$ . Calculate the molar mass of the solute.

(Given vapour pressure of water at $373 K$ is $760 mm Hg$ .)

Accepted Answer

$5 %$ aqueous solution of a non-volatile organic substance by mass means, $100 g$ of solution contains $5 g$ solute and $95 g$ solvent (water).

Thus,

Mass of solute = $W_{b} = 5 g$

Mass of solvent = $W_{a} = 95 g$

Vapour pressure of water at $373 K$ = $p_{a}^{°} = 760 mm$ of $Hg$ .

Vapour pressure of given solution is $= p_{a} = 745 mm$ of $Hg$ .

Molecular mass of solvent (water) = $M_{a} = 18 g$

Molecular mass of solute = $M_{b}$

In a solution the relative lowering of vapour pressure of the solvent is equal to the mole fraction of the non-volatile solute. That is,

$\frac{p_{a}^{0} - p_{a}}{p_{a}^{0}} = Χ_{b}$

$\frac{p_{a}^{0} - p_{a}}{p_{a}^{0}} = \frac{W_{b} \times M_{a}}{W_{a} \times M_{b}}$

On substituting the values we get,

$\frac{760 - 745}{760} = \frac{5.0 \times 18}{95 \times M_{B}} M_{B} = \frac{5.0 \times 18 \times 760}{15 \times 95} M_{B} = 48 {gmol}^{- 1}$

The vapour pressure of a $5 %$ aqueous solution, by mass, of a non-volatile organic substance at $373 K$ is $745 mm$ of $Hg$ . Calculate the molar mass of the solute.

(Given vapour pressure of water at $373 K$ is $760 mm Hg$ .)

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The vapour pressure of a 5%aqueous solution, by mass, of a non-volatile organic substance at 373 K is 745 mm of Hg. Calculate the molar mass of the solute. (Given vapour pressure of water at 373 K is 760 mm Hg.)