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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 $=100^{\circ} \mathrm{C}$ )
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The correct answer is:
$0 \cdot 35 \mathrm{~mol} \mathrm{~kg}^{-1}$
(D)
$\Delta \mathrm{T}_{\mathrm{b}}=(100.18+273)-(100+273)=0.18 \mathrm{~K}$
$\mathrm{K}_{\mathrm{b}}=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \mathrm{~m}=?$
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \quad \therefore \mathrm{m}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{b}}}$
$\therefore \mathrm{m}=\frac{0.18 \mathrm{~K}}{0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}}=0.35 \mathrm{~mol} \mathrm{~kg}^{-1}$
$\Delta \mathrm{T}_{\mathrm{b}}=(100.18+273)-(100+273)=0.18 \mathrm{~K}$
$\mathrm{K}_{\mathrm{b}}=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}, \mathrm{~m}=?$
$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \times \mathrm{m} \quad \therefore \mathrm{m}=\frac{\Delta \mathrm{T}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{b}}}$
$\therefore \mathrm{m}=\frac{0.18 \mathrm{~K}}{0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}}=0.35 \mathrm{~mol} \mathrm{~kg}^{-1}$
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