Question

The relative lowering of vapour pressure of 0.2 molal solution in which the solvent is benzene is

Answer 1

RLVP=n2/n1

=0.2*78/1000

=0.0156

Answer 2

Dear student,

● Answer -
∆P/P = 0.0156

● Explaination -
Mole fraction of solute can be calculated by -
X2 = m × M1 / 1000
X2 = 0.2 × 78 / 1000
X2 = 15.6×10^-3
X2 = 0.0156

Relative lower of vapour pressure is calculated by -
∆P/P = X2
∆P/P = 0.0156

Hence, relative lowering of vapor pressure is 0.0156 .

Thanks for asking...