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Chemistry
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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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 :
JEE Main
JEE Main 2018
Solutions
A
37.5 g
26%
B
75 g
17%
C
150 g
47%
D
50 g
10%
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.
$\frac{75}{100}=\frac{w_{ B } / 50}{w_{ B } / 50+114 / 114}$
$0.75=\frac{w_{ B } / 50}{w_{ B } / 50+1} \Rightarrow w_{ B }=150\, g$