The vapour pressure in mm of Hg,of an aqueous | vedclass

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(D) According to Raoult's law,the relative lowering of vapour pressure is given by: $\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{n_2}{n_1 + n_2}$ where,

Vedclass · Accepted Answer

$752.4$

Vedclass · Answer

(D) According to Raoult's law,the relative lowering of vapour pressure is given by: $\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{n_2}{n_1 + n_2}$ where,$p^{\circ} = 	ext{vapour pressure of pure water at } 100^{\circ} C = 760 \ mmHg$. $p_s = 	ext{vapour pressure of solution at } 100^{\circ} C$. $n_2 = 	ext{moles of glucose} = \frac{18}{180} = 0.1 \ mol$. $n_1 = 	ext{moles of water} = \frac{180}{18} = 10 \ mol$. Substituting these values: $\frac{760 - p_s}{760} = \frac{0.1}{10 + 0.1} = \frac{0.1}{10.1}$. $760 - p_s = 760 	imes \frac{0.1}{10.1} = \frac{76}{10.1} \approx 7.524 \ mmHg$. $p_s = 760 - 7.524 = 752.476 \ mmHg \approx 752.4 \ mmHg$.

The vapour pressure in $mm$ of $Hg$,of an aqueous solution obtained by adding $18 \ g$ of glucose $(C_6H_{12}O_6)$ to $180 \ g$ of water at $100^{\circ} C$ is

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