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Q. 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 :

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Solution:

Relative lowering of vapour pressure is given by

$\frac{\Delta p}{p}=x_{\text {solute }}=\frac{w_{ B } / M_{ B }}{w_{ A } / M_{ A }+w_{ a } / M_{ B }}$

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.

$\frac{75}{100}=\frac{w_{ B } / 50}{w_{ B } / 50+114 / 114}$

$0.75=\frac{w_{ B } / 50}{w_{ B } / 50+1} \Rightarrow w_{ B }=150\, g$