18 g glucose (C6H12O6) is added to 178.2 g water.  The vapor pressure of water (in torr) for this aqueous solution is :

  • Option 1)

     7.6

     

  • Option 2)

    76.0

     

  • Option 3)

    752.4

     

  • Option 4)

    759.0

 

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

 

 Mass of Glucose =\frac{18}{180}=0.1

Moles of Water = =\frac{178.2}{18}=9.9

Mole fraction of Glucose =\frac{1}{100}

From Raoult's law =\frac{760-p_{6}}{760}=\frac{1}{100}

ps = 752.4 tour

Correct  option is 3.

 


Option 1)

 7.6

 

This is an incorrect option.

Option 2)

76.0

 

This is an incorrect option.

Option 3)

752.4

 

This is the correct option.

Option 4)

759.0

This is an incorrect option.

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