Browse by Stream
QnA
Home
Search for Exam, Articles, Questions
QnA
get app
Go To Your Study Dashboard
  1. Home
  2. how
# Engineering

uestion

Asked in: BITSAT-2018

The vapour pressure of water at 20 ^{\circ}C is 17.5 mm  Hg. If 18 g of glucose \left ( C_{6}H_{12}O_{6} \right ) is added to 178.2 g of water at 20 ^{\circ}C the vapour pressure of the resulting solution will be

A.

17.325\: mm \: Hg

B.

17.675\: mm \: Hg

C.

15.750\: mm \: Hg

D.

16.500\: mm \: Hg

PreviousNext

« Previous

Answers (1)
P Pankaj Sanodiya

In a solution containing nonvolatile solute, the pressure  is directly proportional to its mole fraction.

Psolution = vapor pressure of its pure component × mole fraction in solution

P_{\mathrm{sol}}=P^{\circ} X_{\text { solvent }}

Let A be solute and B be solvent

X_{B}=\frac{n_{B}}{n_{A}+n_{B}}=\frac{\frac{178.2}{18}}{\frac{18}{180}+\frac{178.2}{18}}

X_{B}=\frac{9.9}{9.94}=0.99

P_{\text { solution }}=P^{\circ} X_{\text { solvent }}=17.5 \times 0.99

P_{\text { solution }}=17.32.

Similar Questions

Preparation Products

Foundation 2021 Class 10th Maths

Master Maths with "Foundation course for class 10th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Foundation 2021 Class 9th Maths

Master Maths with "Foundation course for class 9th" -AI Enabled Personalized Coaching -200+ Video lectures -Chapter-wise tests.

₹ 350/- ₹ 112/-
Buy Now
Knockout JEE Main April 2021 (One Month)

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

₹ 14000/- ₹ 6999/-
Buy Now
Knockout JEE Main April 2021

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

₹ 22999/- ₹ 14999/-
Buy Now
Test Series JEE Main April 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
Start Now
 
Exams
Articles
Questions