The vapour pressure of water at 20,^{circ}C i | vedclass

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(A) The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^o - P_S}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dil

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$17.325$

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(A) The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^o - P_S}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1}$ for dilute solutions.$1.$ Calculate moles of glucose $(n_2)$: $n_2 = \frac{18 \, g}{180 \, g/mol} = 0.1 \, mol$.$2.$ Calculate moles of water $(n_1)$: $n_1 = \frac{178.2 \, g}{18 \, g/mol} = 9.9 \, mol$.$3.$ Using the formula $\frac{P^o - P_S}{P^o} = \frac{n_2}{n_1 + n_2}$:$\frac{17.5 - P_S}{17.5} = \frac{0.1}{9.9 + 0.1} = \frac{0.1}{10} = 0.01$.$4.$ Solve for $P_S$:$17.5 - P_S = 17.5 	imes 0.01 = 0.175$.$P_S = 17.5 - 0.175 = 17.325 \, mm \, Hg$.

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 $20\,^{\circ}C,$ the vapour pressure of the resulting solution will be $.........\, mm$ of $Hg$.

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