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Two 5 molal solutions are prepared by
dissolving a non-electrolyte non-volatile
solute separately in the solvents X and Y.
The molecular weights of the solvents are
MX and MY , respectively where
M_{X} = \frac{3}{4}M_{Y}. The relative lowering of
vapour pressure of the solution in X is “m”
times that of the solution in Y. Given that
the number of moles of solute is very small
in comparison to that of solvent, the value
of “m” is :

  • Option 1)

    \frac{4}{}3

  • Option 2)

    \frac{3}4

  • Option 3)

    \frac{1}{}2

  • Option 4)

    \frac{1}{}4

 

Answers (1)

 

As we have learnt,

 

Expression of relative lowering of vapour pressure -

\frac{\Delta P}{ P^{0}}= x_{solute}

x_{solute}= \frac{ n_{solute}}{ n_{solute}+n_{solvent}}
 

 

- wherein

\Delta P \: is \: lowering \: o\! f \: v.p.

P^{0}\rightarrow \: vapour\: pressure\: of \: pure\: solvent

x_{solute}\rightarrow \: mole\: fraction \: of \:non\: volatile\: solute

 

 

 

\\*M_{x} = \frac{3}{4} M_{y} \\* \left (\frac{p^{o}- p}{p^{o}} \times \frac{1000}{Mol.\;wt.\;of\;solvent}\right ) = m = 5 \\* \left (\frac{p^{o}_{x}- p_{x}}{p^{o}_{x}} \times \frac{1000}{M_{x}}\right ) = \left (\frac{p^{o}_{y}- p_{y}}{p^{o}_{y}} \times \frac{1000}{M_{y}}\right ) \\* m\left (\frac{p^{o}_{y}- p_{y}}{p^{o}_{y}} \times \frac{1000}{\frac{3}{4}M_{y}}\right ) = \left (\frac{p^{o}_{y}- p_{y}}{p^{o}_{y}} \times \frac{1000}{M_{y}}\right ) \\* m\times\frac{4}{3} = 1 \Rightarrow m = \frac{3}{4}


Option 1)

\frac{4}{}3

Option 2)

\frac{3}4

Option 3)

\frac{1}{}2

Option 4)

\frac{1}{}4

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subam

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