The vapour pressure of benzene at 90,^oC is 1 | vedclass

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(B) Using the relative lowering of vapour pressure formula:$\frac{P^o - P_s}{P_s} = \frac{w 	imes M}{m 	imes W}$Where:$P^o = 1020 \, 	ext{torr}$ (V

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$220$

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(B) Using the relative lowering of vapour pressure formula:$\frac{P^o - P_s}{P_s} = \frac{w 	imes M}{m 	imes W}$Where:$P^o = 1020 \, 	ext{torr}$ (Vapour pressure of pure benzene)$P_s = 990 \, 	ext{torr}$ (Vapour pressure of solution)$w = 5 \, g$ (Mass of solute)$W = 58.5 \, g$ (Mass of solvent,benzene)$M = 78 \, g/mol$ (Molar mass of benzene)$m$ = Molar mass of soluteSubstituting the values:$\frac{1020 - 990}{990} = \frac{5 	imes 78}{m 	imes 58.5}$$\frac{30}{990} = \frac{390}{m 	imes 58.5}$$\frac{1}{33} = \frac{390}{m 	imes 58.5}$$m = \frac{390 	imes 33}{58.5} = 220 \, g/mol$

The vapour pressure of benzene at $90\,^oC$ is $1020 \, \text{torr}$. $5 \, g$ of a solute is dissolved in $58.5 \, g$ of benzene. The vapour pressure of the solution is $990 \, \text{torr}$. What is the molar mass of the solute?

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