The elevation in boiling point of a solution | vedclass

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Question

(A) Step $1$: Calculate the number of moles of $CuCl_2$. $	ext{Moles} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{13.44 \ g}{134.4 \ g \ mol^{-1}

Vedclass · Accepted Answer

$0.16$

Vedclass · Answer

(A) Step $1$: Calculate the number of moles of $CuCl_2$. $	ext{Moles} = \frac{	ext{mass}}{	ext{molar mass}} = \frac{13.44 \ g}{134.4 \ g \ mol^{-1}} = 0.1 \ mol$. Step $2$: Calculate molality $(m)$. $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.1 \ mol}{1 \ kg} = 0.1 \ m$. Step $3$: Determine the van't Hoff factor $(i)$. For $CuCl_2$,the dissociation is $CuCl_2 	o Cu^{2+} + 2Cl^-$. Assuming $100 \%$ ionization,$i = 3$. Step $4$: Calculate elevation in boiling point $(\Delta T_b)$. $\Delta T_b = i 	imes K_b 	imes m = 3 	imes 0.52 	imes 0.1 = 0.156 \ K \approx 0.16 \ K$.

The elevation in boiling point of a solution of $13.44 \ g$ of $CuCl_2$ in $1 \ kg$ of water using the following information will be (Molecular weight of $CuCl_2 = 134.4 \ g \ mol^{-1}$ and $K_b = 0.52 \ K \ kg \ mol^{-1}$)

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