Question

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 to 75%, is :

Answer 1

Answer:

12.5 grams of the solute needs to be dissolved in octane to reduce its vapor pressure to 75%

Explanation:

Given data,

mass of octane, w₁ = 114 g

molar mass of octane(C₈H₁₈), M₁ = 8 x 12 + 18 x 1 = 114 g

mass of solute = w₂

molar mass of solute = M₂ = 50

change in vapor pressure = ΔP/P = 0.25

we know that the change in vapor pressure is related to the weight and mass of the solute as follows;

ΔP/P = (w₂M₁)/(w₁M₂)

=> 0.25 = (w₂ x 114)/(114 x 50)

=> 0.25 = w₂/50

=> w₂ = 0.25 x 50 = 12.5 g

Hence 12.5 grams of the solute needs to be dissolved in octane to reduce its vapor pressure to 75%