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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
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 RegardsArun (askIITians forum expert)
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