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Grade: 12 
        The relative lowering of vapour pressure of an aqueous solution containing non volatile solute is 0.0125. the molality of solution is
10 months ago

Answers : (1)

Arun
14688 Points 
							
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)
10 months ago
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