A solution containing 10 g of an electrolyte | vedclass

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

Question

(B) The dissociation reaction is $AB_2 \rightarrow A^{2+} + 2B^-$. The van't Hoff factor $i$ is given by $i = 1 + (n-1)\alpha$,where $n=3$. So,$i = 1

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) The dissociation reaction is $AB_2 ightarrow A^{2+} + 2B^-$. The van't Hoff factor $i$ is given by $i = 1 + (n-1)\alpha$,where $n=3$. So,$i = 1 + 2\alpha$.The boiling point elevation is $\Delta T_b = T_b - T_b^{\circ} = 100.52^{\circ} C - 100^{\circ} C = 0.52 \ K$.The molality $m$ is calculated as $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{10/200}{100/1000} = \frac{0.05}{0.1} = 0.5 \ mol \ kg^{-1}$.Using the formula $\Delta T_b = i \cdot K_b \cdot m$:$0.52 = (1 + 2\alpha) 	imes 0.52 	imes 0.5$$1 = (1 + 2\alpha) 	imes 0.5$$2 = 1 + 2\alpha$$2\alpha = 1$$\alpha = 0.5 = 5 	imes 10^{-1}$.Thus,the value is $5$.

Similar Questions

Write the value of the van't Hoff factor $(i)$ for $K_{2}SO_{4}$ and $K_{4}[Fe(CN)_{6}]$.

$A$ $0.1 \ m$ aqueous solution of a weak acid $HX$ is $30\%$ ionized. If $K_f$ for water is $1.86 \ ^{\circ}C/m,$ the freezing point of the solution will be ......... $^{\circ}C$.

$A$ $0.1 \ mol$ of a binary electrolyte undergoes complete dissociation in $250 \ g$ of a protic solvent. If the elevation in boiling point is $1.5 \ K$,the molal elevation constant $(K_b)$ is ............. $K \ kg \ mol^{-1}$.

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

The ratio of the value of any colligative property for $KCl$ solution to that for sugar solution is nearly

Vedclass Test Series

Exam Paper Generator

Online Exam Module