The mass of glucose That would be dissolved in 50g of water in order to produce the Same lowering of vapour pressure as in produced by dissolving 1g of urea in the same quantity of water is
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Answer:3g
Explanation:
(Δp)glucose=(Δp)urea
(xB)glucose=(xB)urea
i.e., (nBnA)glucose=(nBnA)urea
wB50×18180=1×1850×60
wB=3g]
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