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$18 \mathrm{~g}$ of glucose $\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)$ dissolved in $1 \mathrm{~kg}$ of water is heated to boiling. The boiling point (in $\mathrm{K}$ ) measured at $1 \mathrm{~atm}$ pressure is closest to [Ebulioscopic constant, $K_{\mathrm{b}}$ for water is $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. Consider absolute zero to be $\left.-273.15^{\circ} \mathrm{C}\right]$
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The correct answer is:
$373.20$
$\begin{aligned} \Delta \mathrm{T}_{\mathrm{b}}=& \mathrm{K}_{\mathrm{b}} \times \mathrm{m} \\=& 0.52 \times \frac{18 / 180}{1} \\ &=0.052 \\ \therefore \mathrm{B} \cdot \mathrm{P} &=373.15+0.052 \\ &=373.2 \mathrm{~K} \end{aligned}$
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