1.

An aqueous solution of urea is found to boil at 100.52^(@)C. Given K_(b) for water is 0.52 K kg mol^(-1), the mole fraction of urea in the solution is

Answer»

1
0.5
0.018
0.25

Solution :`DeltaT_(b)=K_(b)xxm`.
Hence, molality, `m=(DeltaT_(b))/(K_(b))=(0.52)/(0.52)=1`
Molality = 1 means 1 mole of solute in 1000 G of solvent.
But 1000 g of solvent (water)
`=(1000)/(18)" moles "="55.55 moles"`
`therefore"Mole fraction of urea"=(1)/(1+55.55)=0.018`.


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