Q

# Help me solve this - Solutions - JEE Main-3

Liquid $'M'$ and liquid $'N'$ form an ideal solution. The vapour pressures of pure liquids $'M'$ and $'N'$ are $450$ and $700 \; mmHg$, respectively, at the same temperature. Then correct statement is :

$\left ( x_{M}=Mole\; fraction \; of\; 'M' \; in\; solution;$

$x_{N}=Mole\; fraction \; of\; 'N' \; in\; solution;$

$y_{M}=Mole\; fraction \; of\; 'M' \; in\; vapour\; phase;$

$y_{N}=Mole\; fraction \; of\; 'N' \; in\; vapour\; phase )$

• Option 1)

$\frac{x_{M}}{x_{N}}=\frac{y_{M}}{y_{N}}$

• Option 2)

$\left ( x_{M}-y_{M} \right )< \left ( x_{N}-y_{N} \right )$

• Option 3)

$\frac{x_{M}}{x_{N}}< \frac{y_{M}}{y_{N}}$

• Option 4)

$\frac{x_{M}}{x_{N}}> \frac{y_{M}}{y_{N}}$

Views

More volatile component will have greater composition in vapour phase as compared to it's composition in liq.phase

The vapour phase of pure liq. $M$ & $N$ are $450\: mm$ of $Hg$ and $700\: mm$ of $Hg$  respectively

$P^{0}_{N}> P^{0}_{M}$          ($N\rightarrow$ more volatile)

$y_{N}> x_{N}$

$y_{M}< x_{M}\Rightarrow x_{M}> y_{M}$

$y_{N}< x_{M}> x_{N}> y_{M}$

$\frac{x_{M}}{x_{N}}> \frac{y_{M}}{y_{N}}$

Option 1)

$\frac{x_{M}}{x_{N}}=\frac{y_{M}}{y_{N}}$

Option 2)

$\left ( x_{M}-y_{M} \right )< \left ( x_{N}-y_{N} \right )$

Option 3)

$\frac{x_{M}}{x_{N}}< \frac{y_{M}}{y_{N}}$

Option 4)

$\frac{x_{M}}{x_{N}}> \frac{y_{M}}{y_{N}}$

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