Q. 18 g of glucose ($C_{6}H_{12}O_{6}$) is added to 178.2g of water. the vapour pressure of water for this aqueous solution at $100℃$ is

A) 759.00 torr
B) 7.60 torr
C) 76.00 torr
D) 752.40 torr
Explanation

(d)
$\frac{P^{{^\circ}} - P_{s}}{P_{s}} = \frac{n}{N}$
$\ \ \ \ \ \ \ \frac{760 - P_{s}}{P_{s}} = \frac{18/180}{178.2/18}$
=$\frac{0.1}{9.9}$
760-$P_{s} = \frac{1}{99}P_{s}\ \ \ \ \ \ \$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{s} = 752.4$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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