Correct Answer - D
According to depression of freezing point, `Delta T_(f)=K_(f)xx` molality of solution
According to elevation of boiling point,
`Delta T_(b)=K_(b)xx` molality of solution
or `(Delta T_(f))/(Delta T_(b))=(K_(f))/(K_(b))`
Given that
`Delta T_(b)=T_(2)-T_(1)=100.18-100=0.18`
`K_(f)` for water `= 1.86 K kg mol^(-1)`
`K_(b)` for water `= 0.512 K kg mol^(-1)`
`therefore (Delta T_(f))/(0.18)=(1.86)/(0.512)`
`Delta T_(f)=(1.86xx0.18)/(0.512)=0.6539~~0.654`
`Delta T_(f)=T_(1)-T_(2)`
`0.654=0^(@)C-T_(2)`
`therefore T_(2)=-0.654^(@)C`
(`T_(2)rarr` freezing point of aqueous urea solution)