If boiling point of urea solution is $100 \cdot 18^{\circ} \mathrm{C}$ and $\mathrm{kb}$ for water is $0.512…

If boiling point of urea solution is $100 \cdot 18^{\circ} \mathrm{C}$ and $\mathrm{kb}$ for water is $0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, molality of solution is? (Boiling point of water $=100^{\circ} \mathrm{C}$ )
  1. $0 \cdot 25 \mathrm{~mol} \mathrm{~kg}^{-1}$
  2. $0 \cdot 6 \mathrm{~mol} \mathrm{~kg}^{-1}$
  3. $0 \cdot 45 \mathrm{~mol} \mathrm{~kg}^{-1}$
  4. $0 \cdot 35 \mathrm{~mol} \mathrm{~kg}^{-1}$

Solution

$\Delta \mathrm{T}_{\mathrm{b}}=(100.18+273)-(100+273)=0.18 \mathrm{~K}$ $\mathrm{K}_{\mathrm{b}}=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \mathrm{~m}=?$ $\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \quad \therefore \mathrm{m}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{b}}}$ $\therefore \mathrm{m}=\frac{0.18 \mathrm{~K}}{0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}}=0.35 \mathrm{~mol} \mathrm{~kg}^{-1}$

Asked in: MHT CET 2020 (19 Oct Shift 2)

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