The vapour pressure of pure benzene is `639.7 mm Hg` and the vapour pressure of solution of a solute in benzene at the temperature is `631.9 mm Hg`. Calculate the molality of the solution.

Options

A. 0.156 m
B. 1.36 m
C. 0.122 m
D. 0.14 m

Explanation

(1) `x_(B)=(P_("solvent")-P_("solution"))/(P_("solution"))=(639.7-631.9)/(631.9)`
`= 0.012`
0.012 moles are present is 0.988 moles of benzen
`therefore 0.012` moles are present is 77.04 g of benzene
`X_("Solute")=(m)/(m+(1000)/("M.Wt of Solvent")), 0.012 = (m)/(m+12.82)`
`m=0.156 m`