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(C) The relative lowering of vapour pressure is given by the formula: $\frac{\Delta P}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = m 	imes

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$600$

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(C) The relative lowering of vapour pressure is given by the formula: $\frac{\Delta P}{P^o} = \frac{n_2}{n_1 + n_2} \approx \frac{n_2}{n_1} = m 	imes \frac{M_A}{1000}$.Here,$\Delta P = 0.6 \ mm \ Hg$,$m = \frac{1}{18} \ mol \ kg^{-1}$,and $M_A$ (molar mass of water) $= 18 \ g \ mol^{-1}$.Using the relation $\Delta P = P^o 	imes \frac{m 	imes M_A}{1000}$,we find the vapour pressure of pure solvent $P^o = \frac{\Delta P 	imes 1000}{m 	imes M_A} = \frac{0.6 	imes 1000}{(1/18) 	imes 18} = 600 \ mm \ Hg$.Since the solution is dilute,the vapour pressure of the solution $P_s = P^o - \Delta P = 600 - 0.6 = 599.4 \ mm \ Hg$.However,in many textbook contexts,the approximation $\Delta P \approx P_s 	imes \frac{m 	imes M_A}{1000}$ is used,leading to $P_s = \frac{0.6 	imes 1000}{18 	imes (1/18)} = 600 \ mm \ Hg$.

At a certain temperature,when some amount of glucose is dissolved in water,it is observed that the lowering in vapour pressure is $0.6 \ mm \ Hg$ for this solution. What will be the vapour pressure of the same glucose solution if its molality is $(\frac{1}{18}) \ mol \ kg^{-1}$ (in $mm \ Hg$)?

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