The relative lowering of vapour pressure produced by dissolving 71.5 g of substance in 1000g of water is 0.0173. The molecular mass of the substance w
The relative lowering of vapour pressure produced by dissolving 71.5 g of substance in 1000g of water is 0.0173. The molecular mass of the substance will be:
A. 74.39
B. 18
C. 342
D. 60
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Correct Answer - A
`(P_(1)^(@)-P)/P_(1)^(@)=(W_(2) xx M_(1))/(M_(2) xx W_(1))=(71.5 xx 18)/(M_(2) xx 1000)` `:. M_(2)=(71.5 xx 18)/(0.0173 xx 1000)=74.39`
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