0.004 M K_2SO_4 solution is isotonic with 0 | vedclass

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(D) For isotonic solutions, the osmotic pressure $(\pi)$ is equal: $\pi_{K_2SO_4} = \pi_{	ext{glucose}}$.Since $\pi = iCRT$, we have $i_{K_2SO_4} 	i

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(D) For isotonic solutions, the osmotic pressure $(\pi)$ is equal: $\pi_{K_2SO_4} = \pi_{	ext{glucose}}$.Since $\pi = iCRT$, we have $i_{K_2SO_4} 	imes 0.004 	imes RT = 1 	imes 0.01 	imes RT$.Solving for the van't Hoff factor $(i)$: $i = \frac{0.01}{0.004} = 2.5$.For the dissociation of $K_2SO_4$ $(K_2SO_4 ightarrow 2K^+ + SO_4^{2-})$, the number of ions produced $(n)$ is $3$.The degree of dissociation $(\alpha)$ is given by the formula: $i = 1 + (n - 1)\alpha$.Substituting the values: $2.5 = 1 + (3 - 1)\alpha$.$1.5 = 2\alpha$, which gives $\alpha = 0.75$.The percentage dissociation is $\alpha 	imes 100 = 75\%$.

$0.004 \ M \ K_2SO_4$ solution is isotonic with $0.01 \ M$ glucose solution. The percentage dissociation of $K_2SO_4$ is $........$ (Nearest integer).

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