Vapour pressure of a solution containing 6.0gm of a non – volatile solute in 180gm water at $25^{0} C$ is 20 torr. If 1 mole of water is further added, the vapour pressure at $25^{0} C$ increases by 0.02 torr. Identify the correct information

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The molality of final solution is $\frac{1}{1.782} m$

the vapour pressure of pure water at $25^{0} C$ is $\frac{182}{9} torr$

The molality of solution taken initially was $\frac{1}{1.62} m$

The molar mass of solute is 54gm/mol.

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P_{S} = 20 torr \frac{P_{A}^{0} - P_{s}}{P_{s}} = \frac{n_{B}}{n_{A}} = \frac{\frac{6}{M_{B}}}{10} \frac{P_{A}^{0} - 20}{20} = \frac{6}{10 M} (1) \frac{P_{A}^{0} - 20.02}{20.02} = \frac{6}{11 M_{B}} (2) \frac{(P_{A}^{0} - 20) 20.02}{(P_{A}^{0} - 20.02) 20} = \frac{11}{10} \begin{array}{l} 10.01 P_{A}^{0} - 200.2 = 11 (P_{A}^{0} - 20.02) \\ 20.02 = 0.99 P_{A}^{0} \end{array}$
b) $P_{A}^{0} = \frac{20.02}{99} = \frac{182}{9} torr$
$\frac{\frac{182}{9} - 20}{20} = \frac{6}{10} \times \frac{1}{M_{B}} = \frac{2}{9 \times 20}$
a) $M_{B} = \frac{3 \times 9 \times 10}{5} = 54$
c) $m = \frac{\frac{6}{54}}{180} \times 1000 = \frac{100}{9 \times 18} = \frac{1}{1.62} m$
d) $m = \frac{\frac{6}{54}}{198} \times 1000 = \frac{1}{1.782} m$