The vapour pressure of water at $20^{\circ}C\ is\ 17.5\ mmHg.$ $If\ 18\ g\ of\ glucose\ (C_{6}H_{12}O_{6})$ is added to 178.2 g of water at $20^{\circ}C$, the vapour pressure of the resulting solution will be

Question

Dr. Abubacker · Accepted Answer

(d)
Moles of glucose =$\frac{18}{180} = 0.1$
Moles of $H_{2}O = \frac{178.2}{18}\ \ 9.9\$
According to Raoult’s law
$\frac{P^{{^\circ}} - P_{s}}{P^{{^\circ}}} = X_{solute}$
$\ \ \ \frac{17.5 - P_{S}}{17.5\ } = \frac{0.1}{10}\ \ \ \$
so, $P_{s} = 17.325mm\ Hg\$

Q. The vapour pressure of water at $20^{\circ}C\ is\ 17.5\ mmHg.$ $If\ 18\ g\ of\ glucose\ (C_{6}H_{12}O_{6})$ is added to 178.2 g of water at $20^{\circ}C$, the vapour pressure of the resulting solution will be

Explanation

Reviewed by Dr. Abubacker

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