1.

3g of urea is dissolved in 45g of H_(2)O. The relative lowering in vapour pressure is

Answer»

0.05
0.04
0.02
0.01

Solution :Relative lowering of VAPOUR pressure,
`(p_(A)^(@)-p_(A))/(p_(A)^(@))=x_(B)`
where `x_(B)` is the mole fraction of solute (UREA)
`x_("urea")=(n_("urea"))/(n_("urea")+n_(H_(2)O))~~(n_("urea"))/(n_(H_(2)O))` (for dilute solutions)
`n_("urea")=(3)/(60)=0.05`
`n_(H_(2)O)=(45)/(15)=2.5`
`n_("urea")=(0.05)/(2.5)=0.02`


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