If 0.15 g of a solute dissolved in 15 g of | vedclass

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(D) The formula for elevation in boiling point is given by: $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{

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$100$

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(D) The formula for elevation in boiling point is given by: $\Delta T_b = K_b 	imes m$,where $m$ is the molality of the solution.Molality $m = \frac{w 	imes 1000}{M 	imes W}$,where $w$ is the mass of solute,$M$ is the molar mass,and $W$ is the mass of solvent in grams.Substituting the values: $\Delta T_b = \frac{K_b 	imes w 	imes 1000}{M 	imes W}$.Rearranging for $M$: $M = \frac{K_b 	imes w 	imes 1000}{\Delta T_b 	imes W}$.Given: $K_b = 2.16 \ ^oC \ kg \ mol^{-1}$,$w = 0.15 \ g$,$W = 15 \ g$,and $\Delta T_b = 0.216 \ ^oC$.$M = \frac{2.16 	imes 0.15 	imes 1000}{0.216 	imes 15} = \frac{324}{3.24} = 100 \ g \ mol^{-1}$.Thus,the molecular weight is $100$.

If $0.15 \ g$ of a solute dissolved in $15 \ g$ of solvent is boiled at a temperature higher by $0.216 \ ^oC$ than that of the pure solvent,what is the molecular weight of the substance? (Given: molal elevation constant for the solvent is $2.16 \ ^oC \ kg \ mol^{-1}$)

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