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JEE Main 20163 April

Exam: JEE Main
Year: 2016
Date: 3 April
Subject: Chemistry
Chapter: Solutions

JEE Main 2016 · 3 April · Chemistry · Solutions

18g glucose ( C _ 6 H _ 12 O _ 6 ) is added to 178 . 2g water. The...

18g

glucose

\left({C}_{6}{H}_{12}{O}_{6}\right)

is added to

178 . 2g

water. The vapour pressure of water (in torr) for this aqueous solution is:

Options

A $752 . 4$
B $759 . 0$
C $7 . 6$
D $76 . 0$

Correct answer

A. $752 . 4$

Step-by-step solution

\dfrac{∆ P}{{P}^{o}} =

Mole fraction of glucose.

\dfrac{760 - {P}_{Soln}}{760} = \dfrac{\dfrac{{W}_{1}}{Mw{t}_{1}}}{\dfrac{{W}_{1}}{M . w{t}_{1}} + \dfrac{{w}_{2}}{M . w{t}_{2}}} = \dfrac{\dfrac{18}{180}}{\dfrac{18}{180} + \dfrac{178.2}{18}} = \dfrac{0.1}{0.1 + 9.9} = \dfrac{1}{100}

760 - {P}_{Soln} = \dfrac{760}{100}

{P}_{Sol} = 752.4

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