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(B) The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality.Molality $m = \frac{w_2 	imes 1000}{M_2 	ime

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$100$

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(B) The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality.Molality $m = \frac{w_2 	imes 1000}{M_2 	imes w_1}$,where $w_2$ is the mass of solute,$M_2$ is the molar mass of solute,and $w_1$ is the mass of solvent in grams.Given: $\Delta T_b = 0.215 \ K$,$w_2 = 0.15 \ g$,$w_1 = 15 \ g$,$K_b = 2.15 \ K \ kg \ mol^{-1}$.Substituting the values: $0.215 = \frac{2.15 	imes 0.15 	imes 1000}{M_2 	imes 15}$.$0.215 = \frac{2.15 	imes 150}{M_2 	imes 15}$.$0.215 = \frac{2.15 	imes 10}{M_2}$.$M_2 = \frac{21.5}{0.215} = 100 \ g \ mol^{-1}$.

$0.15 \ g$ of a solute is dissolved in $15 \ g$ of a solvent. If the solution boils at a temperature $0.215 \ K$ higher than the boiling point of the pure solvent,what is the molar mass of the solute? $(K_b = 2.15 \ K \ kg \ mol^{-1})$

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