Question

Asked in: BITSAT-2018

18 g glucose (C6H12O6) is added to 178.2 g water.  The vapor pressure of water (in torr) for this aqueous solution is :

A.

 7.6

 

B.

76.0

 

C.

752.4

 

D.

759.0

Answers (1)

The molecular mass of water =2×1+1×16=18g=2×1+1×16=18g

For 178.2g water n_A=9.9

The molecular mass of glucose = 12×1+6×16=180g

For 18g glucose n_B=0.1

X_{B}=\frac{0.1}{0.1+9.9 )}=0.01

X_{A}=0.99

For lowering the vapor pressure,

\begin{aligned} p &=p_{A}^{0} X^{A}=P_{A}^{0}\left(1-X_{B}\right) \\ P &=760(1-0.01) \\ &=760-7.6 \\ &=752.4 \text { torr } \end{aligned}

Preparation Products

Rank Booster NEET 2021

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

₹ 13999/- ₹ 9999/-
Buy Now
Knockout NEET May 2021

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

₹ 22999/- ₹ 14999/-
Buy Now
Knockout NEET May 2022

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

₹ 34999/- ₹ 24999/-
Buy Now
Test Series NEET May 2021

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

₹ 6999/- ₹ 4999/-
Buy Now
Test Series NEET May 2022

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

₹ 6999/-
Buy Now
Exams
Articles
Questions