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The relative lowering of vapour pressure of an aqueous solution containing non volatile solute is 0.0125. the molality of solution is

The relative lowering of vapour pressure of an aqueous solution containing non volatile solute is 0.0125. the molality of solution is

Grade:12

2 Answers

Arun
25750 Points
6 years ago
Dear student
 
Relative lowering of vapour pressure = mole fraction of solute (Roult's law)
p- p(s)/p = x2
 
[p - p(s)]/p = w *M/W*m = 0.0125
 
w/m *W = 0.0125/18 = 0.00070
 
Molality = 0.00070 * 1000
= 0.70 M
 
Regards
Arun (askIITians forum expert)
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Relative lowering of vapour pressure = mole fraction of solute (Roult's law)
p- p(s)/p = x2
[p - p(s)]/p = w *M/W*m = 0.0125
w/m *W = 0.0125/18 = 0.00070
Molality = 0.00070 * 1000
= 0.70 M

Thanks and Regards

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