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 :

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 : (A) 37.5 g (B) 75 g (C) 150 g (D) 50 g

Tardigrade Admin · Accepted Answer

Relative lowering of vapour pressure is given by

\frac{Δ p}{p} = x_{solute} = \frac{w _{B} / M _{B}}{w _{A} / M _{A} + w _{a} / M _{B}}

where

w_{A}

and

w_{B}

are the masses of solvent and solute taken and

M_{A}

and

M_{B}

are the molar masses of the solvent and solute.

\frac{75}{100} = \frac{w _{B} /50}{w _{B} /50 + 114/114}

0.75 = \frac{w _{B} /50}{w _{B} /50 + 1} \Rightarrow w_{B} = 150 g

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

37.5 g

75 g

150 g

50 g

Q. The mass of a non-volatile, non-electrolyte solute (molar mass $= 50 g m o l^{- 1}$ ) needed to be dissolved in $114 g$ octane to reduce its vapour pressure to $75%,$ is :