17.4% (W/V) solution of potassium sulphate ( | vedclass

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

Question

(A) For isotonic solutions,the osmotic pressure is equal: $\pi_{1} = \pi_{2}$,which implies $i_{1} C_{1} RT = i_{2} C_{2} RT$.Concentration $C$ in $mo

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(A) For isotonic solutions,the osmotic pressure is equal: $\pi_{1} = \pi_{2}$,which implies $i_{1} C_{1} RT = i_{2} C_{2} RT$.Concentration $C$ in $mol/L$ is calculated as $\frac{10 	imes \% \ (W/V)}{Molar \ Mass}$.For $K_{2}SO_{4}$: $C_{1} = \frac{10 	imes 17.4}{174} = 1 \ M$.For $NaOH$: $C_{2} = \frac{10 	imes 4}{40} = 1 \ M$. Since $NaOH$ is $100\%$ ionised,$i_{2} = 2$ $(NaOH ightarrow Na^{+} + OH^{-})$.Equating the expressions: $i_{1} 	imes 1 = 2 	imes 1 \implies i_{1} = 2$.For $K_{2}SO_{4} ightarrow 2K^{+} + SO_{4}^{2-}$,the number of ions $n = 3$.The degree of ionisation $\alpha$ is given by $\alpha = \frac{i-1}{n-1} = \frac{2-1}{3-1} = \frac{1}{2} = 0.5$ or $50\%$.

$17.4\% \ (W/V)$ 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 $........... \ \%$.

Similar Questions

The osmotic pressure of a $0.1 \ M$ solution of $NaCl$ and $Na_2SO_4$ will be:

The van't Hoff factor $(i)$ for the reaction of $HNO_3$ with $H_2SO_4$ is:

Van't Hoff factor of $BaCl_2$ in its aqueous solution will be: ($BaCl_2$ is $60 \%$ ionized in the solution)

Which of the following $0.1 \ m$ aqueous solutions exhibits the highest osmotic pressure at $25^{\circ}C$?

In an aqueous solution of potassium ferrocyanide,the degree of dissociation of the salt is $80\%$. The value of the van't Hoff factor is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module