A non-volatile solute ( $0.25$ mol) is dissolved in $4.75$ mol of a volatile solvent whose vapour pressure when pure is $200$ torr. Assuming Raoult's law, the vapour pressure of the solution is

Answer: A

x_{2} = 0.25/ (0.25 + 4.75) = 0.05

and

P = x_{1} P^{0} = 0.95 \times 200 = 190

torr. The value 10 torr is only the lowering

P^{0} - P

. The solution pressure must be below the pure-solvent value of 200 torr.