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?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

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, $M_{2} = 170 g {mole}^{- 1}$