The vapour pressure of water at 20 ^{\circ}C is 17.5 mm  Hg  If 18 g of glucose \left ( C_{6}H_{12}O_{6} \right ) is added to 178.2 g of water at 20 ^{\circ}C the vapour pressure of the resulting solution will be

  • Option 1)

    17.325\: mm \: Hg

  • Option 2)

    17.675\: mm \: Hg

  • Option 3)

    15.750\: mm \: Hg

  • Option 4)

    16.500\: mm \: Hg

 

Answers (1)

As we learnt in

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

 

 p_{solution}= p_{A}^{0}\times \chi_{A}

\chi_{A}=\frac{n_{A}}{n_{B}+n_{A}}=\frac{\frac{178.2}{18}}{\frac{178}{18}+\frac{18}{180}}

\chi_{A}=\frac{9.9}{10}=0.99

p_{solution}=17.5\times 0.99

p_{solution}=17.325

Correct option is 1.

 


Option 1)

17.325\: mm \: Hg

This is the correct option.

Option 2)

17.675\: mm \: Hg

This is an incorrect option.

Option 3)

15.750\: mm \: Hg

This is an incorrect option.

Option 4)

16.500\: mm \: Hg

This is an incorrect option.

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