18 g of glucose C6H12O6 ( molar mass 180 gmol−) is dissolved in 1 kg of water in a sauce pan. The temperature at which water will boil (1.013 bar pressure) Kb for water =0.52 K kg mol− will be

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$18 g$ of glucose $C_{6} H_{12} O_{6}$ ( molar mass $180 g$ ${mol}^{-})$ is dissolved in $1 kg$ of water in a sauce pan. The temperature at which water will boil ( $1.013$ bar pressure) $(K_{b}$ for water $= 0.52 K kg {mol}^{-})$ will be

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$200.202 K$

$373.052 K$

$373.202 K$

$100^{\circ} C$

answer is B.

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Detailed Solution

Weight of solute, $W_{2} = 18 g$

Wt. of solvent, $W_{1} = 1 k g$

$mol . wt . = 180 g {mol}^{-} and K_{b} = 0 .52 K kg {mol}^{-}$

It is known as:

$\begin{array}{l} {ΔT}_{b} = \frac{K_{b} \times W_{2} in g}{M_{2} \times W_{1} in kg} \\ = \frac{0 .52 K kg {mol}^{-} \times 18 g}{180 g {mol}^{-} \times 1 kg} = 0 .052 K \\ T - T_{\circ} = ∆ T_{b} \\ T - 373 K = 0.052 K \\ T = 373 K + 0 .052 K = 373 .052 K \end{array}$