NEET Class neet Answered

Mass of solute = 18 g

Molar mass of benzene = 78.11 g/mol

P = 80% P⁰  = 0.8 P⁰

Relative lowering of vapour pressure = (P⁰ – P) / P⁰ = X_B = Mole fraction of solute

(P⁰ – 0.8 P⁰) / P⁰ = n_B / (n_A + n_B)

0.2 =  n_B / (n_A + n_B)

But number of solute particles is very less compare to solvent particles, hence

0.2  = n_B / n_A

Mole of solvent = 117 / 78.11 ≈ 1.50

Let molar mass of solute be ‘Y’

0.2 = (18 / Y) / 1.50

0.3 = 18/Y

Y = 60 g

Molar mass of solute is 60 g.

In the given answer options, the correct answer is missing.