The vapour pressure of water at 23^{circ} C i | vedclass

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(B) Given: Vapour pressure of pure water $P^{\circ} = 19.8 \ mm$. Moles of solute (glucose) $n_A = 0.1 \ mol$. Moles of solvent (water) $n_B = \frac{1

Vedclass · Accepted Answer

$19.602$

Vedclass · Answer

(B) Given: Vapour pressure of pure water $P^{\circ} = 19.8 \ mm$. Moles of solute (glucose) $n_A = 0.1 \ mol$. Moles of solvent (water) $n_B = \frac{178.2 \ g}{18 \ g/mol} = 9.9 \ mol$. According to Raoult's law for non-volatile solutes: $\frac{P^{\circ} - P_s}{P^{\circ}} = \frac{n_A}{n_A + n_B}$ $\frac{19.8 - P_s}{19.8} = \frac{0.1}{0.1 + 9.9} = \frac{0.1}{10} = 0.01$ $19.8 - P_s = 19.8 	imes 0.01 = 0.198$ $P_s = 19.8 - 0.198 = 19.602 \ mm$.

The vapour pressure of water at $23^{\circ} C$ is $19.8 \ mm$. $0.1 \ mole$ of glucose is dissolved in $178.2 \ g$ of water. What is the vapour pressure (in $mm$) of the resultant solution?

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