The vapour pressure of water at 20^circ C is | vedclass

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Question

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$

Vedclass · Accepted Answer

$17.325$

Vedclass · Answer

(D) The relative lowering of vapour pressure is given by the formula: $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ (for dilute solutions).Here,$P^\circ = 17.5 \ mm \ Hg$,$W_{solute} = 18 \ g$,$M_{solute} = 180 \ g/mol$,$W_{solvent} = 178.2 \ g$,$M_{solvent} = 18 \ g/mol$.Moles of glucose $(n_2) = \frac{18}{180} = 0.1 \ mol$.Moles of water $(n_1) = \frac{178.2}{18} = 9.9 \ mol$.Using the formula $\frac{P^\circ - P_s}{P^\circ} = \frac{n_2}{n_1 + n_2}$:$\frac{17.5 - P_s}{17.5} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$.$17.5 - P_s = 17.5 	imes 0.01 = 0.175$.$P_s = 17.5 - 0.175 = 17.325 \ mm \ Hg$.

The vapour pressure of water at $20^\circ C$ is $17.5 \ mm \ Hg$. If $18 \ g$ of glucose $(C_6H_{12}O_6)$ is added to $178.2 \ g$ of water at $20^\circ C$,what will be the vapour pressure of the resulting solution (in $mm \ Hg$)?

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