The vapour pressure of an aqueous solution of sugarcane molecular weight 342 is 756 mm Hg at 100-C- How many grams of sugar are present per 1000 g of water- Vapour pressure of pure water at 100-C is 760 mm Hg

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Relative Lowering of Vapour Pressure

The vapour pr...

Question

The vapour pressure of an aqueous solution of sugarcane (molecular weight 342) is $756 m m H g at 100^{\circ} C$ . How many grams of sugar are present per $1000 g$ of water?
Vapour pressure of pure water at $100^{\circ} C$ is $760 m m H g$

50 g

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77.5 g

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100 g

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110 g

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Solution

The correct option is C $100 g$
When a non volatile solute is added to a volatile solvent, the vapour pressure of the solution is lowered as compared to pure solvent.
For dilute solutions, relative lowering of vapour pressure is given by :
$\frac{Δ P}{P_{s o l v e n t}^{\circ}} = χ_{s o l u t e}$

$\frac{Δ P}{P_{s o l v e n t}^{\circ}} = \frac{n_{s o l u t e}}{n_{s o l v e n t} + n_{s o l u t e}}$

For dilute solutions :
$n_{s o l v e n t} >> n_{s o l u t e}$

$\frac{Δ P}{P_{s o l v e n t}^{\circ}} = \frac{n_{s o l u t e}}{n_{s o l v e n t}}$

$\Rightarrow \frac{n_{s o l u t e}}{n_{s o l v e n t}} = \frac{760 - 756}{760}$

$\Rightarrow \frac{\frac{weight of solute}{342}}{\frac{1000}{18}} = \frac{4}{760} Weight of solute = \frac{4 \times 342 \times 1000}{760 \times 18} = 100 g$

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Relative Lowering of Vapour Pressure

Standard XII Chemistry

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The vapour pressure of an aqueous solution of sugarcane (molecular weight 342) is 756 mm Hg at 100∘C. How many grams of sugar are present per 1000 g of water? Vapour pressure of pure water at 100∘C is 760 mm Hg

The vapour pressure of an aqueous solution of sugarcane (molecular weight 342) is $756 m m H g at 100^{\circ} C$ . How many grams of sugar are present per $1000 g$ of water?
Vapour pressure of pure water at $100^{\circ} C$ is $760 m m H g$