When 0.01 mol of nonvolatile solute is disso | vedclass

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Question

(C) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} 	imes m$,where $m$ is the molality of the solution. Molality $m = \frac{	ext

Vedclass · Accepted Answer

$0.033$

Vedclass · Answer

(C) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} 	imes m$,where $m$ is the molality of the solution. Molality $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in } kg}$. Given: $\Delta T_{b} = 0.6 \ K$,$K_{b} = 2 \ K \ kg \ mol^{-1}$,and moles of solute $= 0.01 \ mol$. Substituting the values: $0.6 = 2 	imes \frac{0.01}{W_{solvent}}$. $W_{solvent} = \frac{2 	imes 0.01}{0.6} = \frac{0.02}{0.6} = \frac{1}{30} \ kg$. $W_{solvent} \approx 0.0333 \ kg$. Thus,the correct option is $C$.

When $0.01 \ mol$ of nonvolatile solute is dissolved in a certain solvent,calculate the mass of the solvent in $kg$ if $\Delta T_{b} = 0.6 \ K$ and $K_{b}$ for the solvent $= 2 \ K \ kg \ mol^{-1}$. (in $kg$)

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