Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

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.

Posted by

perimeter

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Result 2024 Live: JEE Mains session 2 results at jeemain.nta.ac.in soon; rank predictor
anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores
anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question