Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Solve it, 18 g glucose (C6H12O6) is added to 178.2 g water. The vapor pressure of water (in torr) for this aqueous solution is :

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)

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

 

 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.

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 2025 chemistry needs conceptual understanding; 5 tips to score high
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon; paper pattern, exams from January 22
anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters

Latest Question