Vapour pressure of water at 293 K is 17.535 ...

NCERT Class XII Chemistry
Chapter - Solutions
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Question : 34 of 53

Marks: +1, -0

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g water.

Solution:

Given,

P^{0} = 17.535

mm Hg,

w_2 = 25\ \mathrm{g}

w_1 = 450\ \mathrm{g}

P_s = ?

For solute (glucose,

\mathrm{C_6H_{12}O_6}

M_2 = 180\ \mathrm{g\ mol^{-1}}

For solvent (

\mathrm{H_2O}

M_1 = 18

g mol

^{-1}

Applying Raoult’s law,

\frac{P^{0}-P_{s}}{P^{0}} = \frac{n_{2}}{n_{1}+n_{2}}

\frac{P^{0}-P_{s}}{P^{0}} = \frac{n_{2}}{n_{1}} = \frac{ \frac{w_{2}}{M_{2}} }{ \frac{w_{1}}{M_{1}} }

or,

1 - \frac{P_{s}}{P^{0}} = \frac{w_{2} M_{1}}{w_{1} M_{2}}

(\because n_2 \ll n_1)

Substituting the given value, we get

1 - \frac{P_{s}}{17.535} = \frac{25 \times 18}{450 \times 180}

or,

1 - \frac{P_{s}}{17.535} = \frac{1}{180}

1 - \frac{1}{180} = \frac{P_{s}}{17.535}

or,

\frac{179}{180} = \frac{P_{s}}{17.535}

or,

P_{s} = 17.535 \times \frac{179}{180}

= 17.437\ \mathrm{mm\ Hg}

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