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Home  >>  AIMS  >>  Class12  >>  Chemistry  >>  Liquid Solutions
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The vapour pressure of pure benzene of a certain temperature is 0.850 bar. A non volatile, non-electrolyte solid weighing 0.5g is added to 39.0 g of benzene (molar mass $78g mol^{−1}$). The vapour pressure of the solution then is 0.845 bar. What is the molar mass of the solid substance?

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$W_2=0.5g,W_1=39.0g,M_1=78gmol^{-1}$
$P^0=0.850$bar
$P^S=0.845$bar
Substituting these values in the formula,
$\large\frac{P^0-P_1}{P^0}=\frac{n_2}{n_1}=\frac{W_2/M_2}{W_1/M_1}=\frac{W_2M_1}{W_1M_2}$
we get $\large\frac{0.850bar-0.845bar}{0.850bar}=\frac{0.5g\times 78gmol^{-1}}{39.0g\times M_2}$
Or $M_2=\large\frac{0.5\times 78}{39.0}\times \frac{0.850}{0.005}$$gmol^{-1}=170gmol^{-1}$
Hence (A) is the correct answer.
answered Jun 10, 2014 by sreemathi.v
 

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