18 , g of glucose (C_{6}H_{12}O_{6}) dissolve | vedclass

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Question

(C) Given,mass of glucose $(C_{6}H_{12}O_{6})$,$w_{2} = 18 \, g$. Mass of solvent (water),$w_{1} = 1 \, kg$. $K_{b}$ of water $= 0.52 \, K \, kg \, mo

Vedclass · Accepted Answer

$373.20$

Vedclass · Answer

(C) Given,mass of glucose $(C_{6}H_{12}O_{6})$,$w_{2} = 18 \, g$. Mass of solvent (water),$w_{1} = 1 \, kg$. $K_{b}$ of water $= 0.52 \, K \, kg \, mol^{-1}$. Molar mass of glucose $(M_{2}) = 6 	imes 12 + 12 	imes 1 + 6 	imes 16 = 180 \, g \, mol^{-1}$. Number of moles of glucose $(n_{2}) = \frac{18 \, g}{180 \, g \, mol^{-1}} = 0.1 \, mol$. Molality $(m) = \frac{n_{2}}{w_{1} 	ext{ (in kg)}} = \frac{0.1 \, mol}{1 \, kg} = 0.1 \, m$. Elevation in boiling point,$\Delta T_{b} = K_{b} 	imes m = 0.52 	imes 0.1 = 0.052 \, K$. Boiling point of solution,$T_{b} = T_{b}^{\circ} + \Delta T_{b} = 373.15 \, K + 0.052 \, K = 373.202 \, K$. Thus,the boiling point is closest to $373.20 \, K$.

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