1.

The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular weight of the substance will be

Answer»

<P>`18.0`
342
60
180

Solution :`(P^(@)-P_(s))/(P^(@))=((W)/(m))/((w)/(m)+(W)/(M))`
or `0.00713 = (71.5//m)/((71.5)/(m)+(1000)/(18))`


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