A solution containing 0.85 g of ZnCl_2 in 12 | vedclass

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

Question

(B) The molar mass of $ZnCl_2$ is $65.3 + 2 	imes 35.5 = 136.3 \ g \ mol^{-1}$.The observed molar mass $(M_{obs})$ is calculated using the depression

Vedclass · Accepted Answer

$73.5 \ \%$

Vedclass · Answer

(B) The molar mass of $ZnCl_2$ is $65.3 + 2 	imes 35.5 = 136.3 \ g \ mol^{-1}$.The observed molar mass $(M_{obs})$ is calculated using the depression in freezing point formula:$\Delta T_f = K_f 	imes \frac{w 	imes 1000}{M_{obs} 	imes W}$$0.23 = 1.86 	imes \frac{0.85 	imes 1000}{M_{obs} 	imes 125}$$M_{obs} = \frac{1.86 	imes 0.85 	imes 1000}{0.23 	imes 125} \approx 55.0 \ g \ mol^{-1}$.The Van't Hoff factor $(i)$ is given by:$i = \frac{M_{normal}}{M_{obs}} = \frac{136.3}{55.0} \approx 2.478$.For the dissociation $ZnCl_2 ightleftharpoons Zn^{2+} + 2Cl^-$,the Van't Hoff factor is $i = 1 + (n-1)\alpha$,where $n=3$.$i = 1 + 2\alpha$.$2.478 = 1 + 2\alpha \implies 2\alpha = 1.478 \implies \alpha = 0.739$.Thus,the degree of dissociation is approximately $73.5 \ \%$.

$A$ solution containing $0.85 \ g$ of $ZnCl_2$ in $125.0 \ g$ of water freezes at $-0.23 \ ^\circ C$. The apparent degree of dissociation of the salt is ($K_f$ for water $= 1.86 \ K \ kg \ mol^{-1}$,atomic mass: $Zn = 65.3$ and $Cl = 35.5$)

Similar Questions

Elevation in boiling point of the aqueous solution of $0.01 \ M \ BaCl_2$ compared to $0.01 \ M$ urea is . . . . . . .

For a $0.25 \ m$ solution of $SrCl_2$,the van't Hoff factor is found to be $2.5$. What will be the effective molality of the solution in $m$?

Which of the following has the same van't Hoff factor $(i)$ as $K_3[Fe(CN)_6]$?

The freezing point of a $1.00 \ m$ aqueous solution of $HF$ is found to be $-1.91 \ ^oC$. The freezing point constant of water,$K_f$ is $1.86 \ K \ kg \ mol^{-1}$. The percentage dissociation of $HF$ at this concentration is ................ $\%$

At $25 \ ^oC$,the highest osmotic pressure is exhibited by $0.1 \ M$ solution of

Vedclass Test Series

Exam Paper Generator

Online Exam Module