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)

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.

Preparation Products

JEE Main Rank Booster 2021

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout JEE Main April 2021 (Subscription)

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 6999/- ₹ 4999/-
Buy Now
Knockout JEE Main April 2022 (Subscription)

Knockout JEE Main April 2022 Subscription.

₹ 8999/- ₹ 5999/-
Buy Now
Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 22999/- ₹ 14999/-
Buy Now
Knockout JEE Main April 2022

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 34999/- ₹ 24999/-
Buy Now
Exams
Articles
Questions