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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 :

  • Option 1)

    37.5 g

  • Option 2)

    75 g

  • Option 3)

    150 g

  • Option 4)

    50 g

 

Answers (1)

best_answer

As we have learned

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

 

\frac{p^{0}-p^{s}}{p^{0}}=\frac{n}{n+N}

\frac{100-75}{100}=\frac{n}{n+\frac{114}{114}}=\frac{n}{n+1}

25 n + 25 = 100n

75m=25

n=1/3 = wt/50

wt= 50/3

 

 

 

 

 

 


Option 1)

37.5 g

This is incorrect

Option 2)

75 g

This is incorrect

Option 3)

150 g

This is incorrect

Option 4)

50 g

This is incorrect

Posted by

prateek

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