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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}}

 
Answers (1)
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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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