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Q.
The ionic strength of a solution containing 0.1 mole/kg of KCl and 0.2 mole/kg of $CuSO_4$ is
Odisha JEEOdisha JEE 2003Solutions
Solution:
Ionic strength is half of the sum of all the terms obtained by multiplying the molality of each ion by the square of its valency
$I = \frac{1}{2} (m_1z_1^2 + m_2z^2_2 + m_2z_2^2+m_3z^2_3 + m_4z^2_4)$
Here for $K^+$ ion,
$m_1 = 0.1 \, m , z_1 = 1$
For $Cl^{-1}$ ion,
$m_2 = 0.1 \, m , z_2 = 1$
For $Cu^{2+} $ ion
$m_3 = 0.2 \, m, z_3 =2 $
For $SO^{2-}_4 $ ion
$m_4 = 0.2 \, m ,z_4 = 2$
$\therefore $ $I = \frac{1}{2} [0.1 \times 1^2 + 0.1 \times 1^2 + 0.2 \times 2^2 + 0.2 \times 2^2]$
= $\frac{1}{2} [ 0.1 + 0.1 + 0.8 + 0.8 ] $
= $\frac{18}{z}$ = 0.9