The vapour pressure of `C Cl_4 at 25^@C` is 143 mm Hg, if 0.5 g of a non-volatile solute (mol. weight = 65) is dissolved in 100 g of `C CI_4`, the vapour pressure of the solution will be

Options

A. 94.39 mm Hg
B. 199.34 mm Hg
C. 141.31 mm Hg
D. 143.99 mm Hg

Answer

C

Explanation

`p^@ = 143` mm = vapour pressure of pure solvent
`P_s` = vapour pressure of solute
`w_A`= weight of solute = 0.5 g
`w_B` = weight of solvent = 100
`M_A` = molar mass of solute = 65
`M_B`=molar mass of solvent = 154
`(P^@ – P_s)/(P^@) = w_A/M_A xx M_s/w_B implies (143 – P_s)/143 = 0.5/65 xx 154/100`
`implies P_s= 141.31 mm Hg`