Calculate the number of moles of nonvolatile | vedclass

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(B) 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 $(m)$ i

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$0.2$

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(B) 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 $(m)$ is defined as the number of moles of solute $(n_{solute})$ divided by the mass of the solvent in kilograms $(W_{solvent(kg)})$.So,$\Delta T_{b} = K_{b} 	imes \frac{n_{solute}}{W_{solvent(kg)}}$.Given: $\Delta T_{b} = 0.8 \ K$,$K_{b} = 2 \ K \ kg \ mol^{-1}$,and $W_{solvent(kg)} = 0.5 \ kg$.Substituting the values: $0.8 = 2 	imes \frac{n_{solute}}{0.5}$.$0.8 = 4 	imes n_{solute}$.$n_{solute} = \frac{0.8}{4} = 0.2 \ mol$.

Calculate the number of moles of nonvolatile solute dissolved in $0.5 \ kg$ solvent if molal elevation constant for solvent is $2 \ K \ kg \ mol^{-1}$ and the elevation in boiling point is $\Delta T_{b} = 0.8 \ K$.

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