If a 0.02 , M , Pb(NO_3)_2 solution is isoton | vedclass

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(C) For isotonic solutions,the osmotic pressure is equal,so $\pi_1 = \pi_2$. Since $\pi = iCRT$,and $T$ is constant,we have $i_1 C_1 = i_2 C_2$. For g

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$0.75$

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(C) For isotonic solutions,the osmotic pressure is equal,so $\pi_1 = \pi_2$. Since $\pi = iCRT$,and $T$ is constant,we have $i_1 C_1 = i_2 C_2$. For glucose,$i_2 = 1$ (non-electrolyte),so $i_1 	imes 0.02 = 1 	imes 0.05$. $i_1 = \frac{0.05}{0.02} = 2.5$. For $Pb(NO_3)_2 ightarrow Pb^{2+} + 2NO_3^-$,the number of ions $n = 3$. The van't Hoff factor $i = 1 + \alpha(n - 1)$. $2.5 = 1 + \alpha(3 - 1)$. $2.5 = 1 + 2\alpha$. $1.5 = 2\alpha$. $\alpha = 0.75$.

If a $0.02 \, M \, Pb(NO_3)_2$ solution is isotonic with a $0.05 \, M$ aqueous glucose solution,the degree of dissociation of $Pb(NO_3)_2$ will be ..........

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