Question 12 Marks
The vapour pressure of pure benzene at a certain temperature is $0.850$ bar. A non-volatile, non-electrolyte solid weighing $0.5\ g$ when added to $39.0\ g$ of benzene $($molar mass $78\ g\ mol^{-1}).$ Vapour pressure of the solution, then, is $0.845$ bar. What is the molar mass of the solid substance?
Answer
View full question & answer→The various quantities known to us are as follows:
$p _1^0=0.850$ bar; $p =0.845$ bar; $M _1=78\ g\ mol ^{-1} ; w _2=0.5 g ; w _1=39 g$
Substituting these values in equation of relative lowering of vapour pressure, we get
$\frac{0.850 bar -0.845 bar }{0.850 bar }=\frac{0.5 g \times 78\ g\ mol ^{-1}}{M_2 \times 39\ g}$
Therefore, $M_{2 }= 170\ g\ mol^{-1}$
$p _1^0=0.850$ bar; $p =0.845$ bar; $M _1=78\ g\ mol ^{-1} ; w _2=0.5 g ; w _1=39 g$
Substituting these values in equation of relative lowering of vapour pressure, we get
$\frac{0.850 bar -0.845 bar }{0.850 bar }=\frac{0.5 g \times 78\ g\ mol ^{-1}}{M_2 \times 39\ g}$
Therefore, $M_{2 }= 170\ g\ mol^{-1}$