The rise in boiling point of a solution conta | vedclass

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(A) Given: Mass of glucose $(w) = 1.8 \ g$ Mass of solvent $(W) = 100 \ g$ Elevation in boiling point $(\Delta T_b) = 0.1^{\circ} C$ Molar mass of glu

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$1 \ K \ kg / mol$

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(A) Given: Mass of glucose $(w) = 1.8 \ g$ Mass of solvent $(W) = 100 \ g$ Elevation in boiling point $(\Delta T_b) = 0.1^{\circ} C$ Molar mass of glucose $(M) = 180 \ g / mol$ The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is molality. Molality $(m) = \frac{w 	imes 1000}{M 	imes W} = \frac{1.8 	imes 1000}{180 	imes 100} = 0.1 \ mol / kg$ Now,$K_b = \frac{\Delta T_b}{m} = \frac{0.1}{0.1} = 1 \ K \ kg / mol$.

The rise in boiling point of a solution containing $1.8 \ g$ of glucose in $100 \ g$ of solvent is $0.1^{\circ} C$. The molal elevation constant of the liquid is

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