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(A) The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality.The molality $m$ is given by $\frac{w 	imes 1

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(A) The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality.The molality $m$ is given by $\frac{w 	imes 1000}{M_2 	imes W}$,where $w$ is the mass of solute,$M_2$ is the molar mass of solute,and $W$ is the mass of solvent in grams.Given: $\Delta T_b = 0.1 \ ^\circ C$,$w = 1.8 \ g$,$M_2 = 180 \ g/mol$ (for glucose),$W = 100 \ g$.Substituting the values: $0.1 = K_b 	imes \frac{1.8 	imes 1000}{180 	imes 100}$.$0.1 = K_b 	imes \frac{1800}{18000} = K_b 	imes 0.1$.Therefore,$K_b = \frac{0.1}{0.1} = 1 \ K \ kg/mol$.

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

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