Question

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

Answer 1

Answer:

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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

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