If 13.44 g of CuCl_2 is dissolved in 1 kg o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The dissociation of $CuCl_2$ is: $CuCl_2 \rightarrow Cu^{2+} + 2Cl^-$. Since there are $3$ ions produced,the van't Hoff factor $i = 3$ (assuming c

Vedclass · Accepted Answer

$0.16$

Vedclass · Answer

(A) The dissociation of $CuCl_2$ is: $CuCl_2 ightarrow Cu^{2+} + 2Cl^-$. Since there are $3$ ions produced,the van't Hoff factor $i = 3$ (assuming complete dissociation). The molality $m$ is calculated as: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{13.44 \ g / 134.1 \ g \ mol^{-1}}{1 \ kg} \approx 0.1 \ mol \ kg^{-1}$. The elevation in boiling point is given by $\Delta T_b = i 	imes K_b 	imes m$. $\Delta T_b = 3 	imes 0.5 	imes 0.1 = 0.15 \ K$. Rounding to the nearest provided option,the value is $0.16$.

If $13.44 \ g$ of $CuCl_2$ is dissolved in $1 \ kg$ of water,calculate the elevation in boiling point of the solution. $[K_b = 0.5 \ K \ kg \ mol^{-1}$,molar mass of $CuCl_2 = 134.1 \ g \ mol^{-1}]$

Similar Questions

Which of the following will show maximum osmotic pressure for its $1 \, m$ solution?

$A$ $0.2 \ m$ solution of acid $HX$ is $20\%$ ionized in the solution. Given $K_f = 1.86 \ K \ kg \ mol^{-1}$,the freezing point of the solution is ........... $^oC$.

In which case is Raoult's law not applicable?

The van't Hoff factor $(i)$ for a dilute aqueous solution of the strong electrolyte barium hydroxide is

$17.4\% \ (W/V)$ solution of potassium sulphate (mol. $wt. = 174$) is isotonic with $4\% \ (W/V)$ aqueous solution of $NaOH$. If $NaOH$ is $100\%$ ionised,the degree of ionisation of potassium sulphate is $........... \ \%$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module