# The vapour pressure of pure benzene is 640mm og Hg. 2.175×10-3kg of non-vloatile solute is added to 39 gram of benzene the vapour pressure of solution is 600mm of HG. - Chemistry

The vapour pressure of pure benzene is 640mm og Hg. 2.175×10-3kg of non-vloatile solute is added to 39 gram of benzene the vapour pressure of solution is 600mm of HG. Calculate molar mass of solute.

[C = 12, H = 1]

#### Solution

Molar mass of benzene C6H6 = (6 × 12 + 6 × 1) × 10-3 kg mol-1

p10 = 640mm Hg, p = 600mm Hg

W1 = 39 × 10-3 kg

W2 = 2.175 × 10-3 kg mol-1

M1 = 78 × 10-3 kg mol-1

M2 = ?

(p_1^0-p)/p_1^0 = W_2/M_2 M_1/W_1

(640mm-600mm)/(640mm) = (2.175xx10^-3"kg"xx78.0xx10^-3"kg mol"^-1)/(39.0xx10^-3"kg"xxM_2)

M_2 = (2.175xx10^-3"kg"xx78.0xx10^-3"kg mol"^-1xx640"mm")/(39.0xx10^-3"kg"xx40"mm")

M_2 = 69.6xx10^-3"kg mol"^-1

"Molecular mass" = 69.6"g mol"^-1

Concept: Colligative Properties and Determination of Molar Mass - Introduction
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