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18 g of glucose \left ( C_{6}H_{12}O_{6} \right ) is added to 178.2 g of water. The vapour pressure of water for this aqueous solution at 100^{\circ}C is

  • Option 1)

    759.00 torr

  • Option 2)

    7.60 torr

  • Option 3)

    76.00 torr

  • Option 4)

    752.40 torr.

 

Answers (1)

best_answer

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

 

 \frac{p_{A}^{0}-p_{s}}{p_{A}^{0}}=\chi_{B}

Mole fraction of glucose(\chi_{B})=\frac{18/180}{\frac{18}{180}+\frac{178.2}{18}}=\frac{1}{100}

p_{A}^{0}= Vapour pressure of pure water at 100 co = 760 torr

\frac{760-p_{s}}{760}=\frac{1}{100}

ps = Vapour pressure of solution = 752.4 torr.

Correct option is 4.

 


Option 1)

759.00 torr

This is an incorrect option.

Option 2)

7.60 torr

This is an incorrect option.

Option 3)

76.00 torr

This is an incorrect option.

Option 4)

752.40 torr.

This is the correct option.

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perimeter

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