The mass of a non-volatile, non-electrolyte solute (molar mass = 50 g mol - 1 ) needed to be dissolved in 114 g octane to reduce its vapour pressure by 75 %, is:

Step-by-step solution

Molar mass of octane = 114 g/mol. From the lowering of vapour pressure, we have, ∆ P P = W 2 M 2 W 2 M 2 + W 1 M 1 Where W 2   a n d   M 2 are mass and molar mass of solute and W 1   a n d   M 1 are mass and molar mass of octane. 75 100 = W 2 50   g / m o l W 2 50   g / m o l + 114   g 114   g / m o l   0.75 = W 2 50 W 2 50 + 1 W 2 50 + 1 = W 2 50 × 0.75 W 2 = 150 g