180 g of glucose,C_6H_{12}O_6,is dissolved i | vedclass

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(A) The elevation in boiling point is given by the formula: $\Delta T_b = i 	imes K_b 	imes m$. Since glucose is a non-electrolyte,the van't Hoff fa

Vedclass · Accepted Answer

$373.67$

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(A) The elevation in boiling point is given by the formula: $\Delta T_b = i 	imes K_b 	imes m$. Since glucose is a non-electrolyte,the van't Hoff factor $i = 1$. The molality $m$ is calculated as: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{180 \ g / 180 \ g \ mol^{-1}}{1 \ kg} = 1 \ mol \ kg^{-1}$. Substituting the values: $\Delta T_b = 1 	imes 0.52 \ K \ kg \ mol^{-1} 	imes 1 \ mol \ kg^{-1} = 0.52 \ K$. The boiling point of the solution $T_b$ is: $T_b = T_b^0 + \Delta T_b = 373.15 \ K + 0.52 \ K = 373.67 \ K$.

$180 \ g$ of glucose,$C_6H_{12}O_6$,is dissolved in $1 \ kg$ of water in a vessel. The temperature at which water boils at $1.013 \ bar$ is $ . . . . . . $ (given,$K_b$ for water is $0.52 \ K \ kg \ mol^{-1}$. Boiling point for pure water is $373.15 \ K$) (in $K$)

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