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
Options:
A .  
180
B .  
342
C .  
60
D .  
18.0
Answer: Option A : A △P1P01=n2n1+n2≈n2n1=W2/M2W1/M1 M2=W2×M1W1×(△P1P01)=71.5×181000×0.00713=180.5g/mole
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