Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002. The molality of the solution is

Question

Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002. The molality of the solution is (A) 0.11 (B) 0.021 (C) 0.004 (D) 0.222

Tardigrade Admin · Accepted Answer

(P0-P/P0)=(W2/M2)×(M1/W1) 0.002=(W2/M2)×(18/1000) (w2/M2)=0.11 mole Molality=(w2/M2)×(1000/W1)=0.11×(1000/1000)=0.11 m

Q. Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002. The molality of the solution is

0.11

0.021

0.004

0.222

$\frac{P _{0} - P}{P _{0}} = \frac{W _{2}}{M _{2}} \times \frac{M _{1}}{W _{1}}$
$0.002 = \frac{W _{2}}{M _{2}} \times \frac{18}{1000}$
$\frac{w _{2}}{M _{2}} = 0.11 m o l e$
$M o l a l i t y = \frac{w _{2}}{M _{2}} \times \frac{1000}{W _{1}} = 0.11 \times \frac{1000}{1000} = 0.11 m$

Q. Relative lowering of vapour pressure of a dilute solution of glucose dissolved in 1 kg of water is 0.002. The molality of the solution is

0.11

0.021

0.004

0.222

P0​P0​−P​=M2​W2​​×W1​M1​​ 0.002=M2​W2​​×100018​ M2​w2​​=0.11mole Molality=M2​w2​​×W1​1000​=0.11×10001000​=0.11m

$\frac{P _{0} - P}{P _{0}} = \frac{W _{2}}{M _{2}} \times \frac{M _{1}}{W _{1}}$
$0.002 = \frac{W _{2}}{M _{2}} \times \frac{18}{1000}$
$\frac{w _{2}}{M _{2}} = 0.11 m o l e$
$M o l a l i t y = \frac{w _{2}}{M _{2}} \times \frac{1000}{W _{1}} = 0.11 \times \frac{1000}{1000} = 0.11 m$