13.44 g of CuCl_2 is dissolved in 1 kg of w | vedclass

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Question

(A) Step $1$: Calculate the number of moles of $CuCl_2$: $n = \frac{13.44 \ g}{134.4 \ g \ mol^{-1}} = 0.1 \ mol$. Step $2$: Calculate molality $(m)$:

Vedclass · Accepted Answer

$0.16$

Vedclass · Answer

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

$13.44 \ g$ of $CuCl_2$ is dissolved in $1 \ kg$ of water. Determine the elevation in boiling point of the solution. [$K_b = 0.52 \ K \ kg \ mol^{-1}$,$mol \ wt$ of $CuCl_2 = 134.4$]

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