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Can someone help me with this, - Solutions - JEE Main

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)
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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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