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Q.

The vapour pressure of pure benzene at a certain temperature is 0.850bar, A non - volatile, non-electrolyte solid weighing 0.5g when added to 39.0g of benzene (molar mass 78g mol^-1). vapour pressure of the solution then, is 0.845bar. what is the molar mass of the solid substance?

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Detailed Solution

$P^{0} =$ V.P of pure benzene = 0.850 bar
$P_{s}$ = V.P of solution = 0.845 bar
$M_{1}$ = M wt of solvent = 78g mole^-1
$ω_{1}$ = wt of solvent = 39g
$ω_{2}$ = wt of solute = 0.5g
Substituting these values in equation
$\frac{P^{0} - P_{S}}{P^{0}} = \frac{ω_{2}}{M_{2}} \times \frac{M_{1}}{ω_{1}}$
we get $\frac{0 .0850 bar - 0 .845 bar}{0 .850 bar}$
$= \frac{0.5 g \times 78 g {mol}^{- 1}}{M_{2} \times 39 g}$
There fore, $M_{2} = 170 g {mole}^{- 1}$

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