Question

25 ° C. An aqueous solution of iodine containing 0.0516 g liter at temperature 4: 412 g liter containing CCL, is in equilibrium with the solution. 25 C If the solubility of iodine in water at temperature is 0.34 g liter-1, multiply its solubility in cc.

Answer 1

First we find the value of K:

K = $\dfrac{C_{CCl_4}}{C_{H_2O}} = \frac{4.412}{0.0516} = 85.5$

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To find the solubility of iodine in CCl4 we should use a distribution law:

$\dfrac{Solubility \ of \ iodine \ in \ CCl_4}{Solubility \ of \ iodine \ in \ water}=85.5$

Solubility of iodine in CCl4 = $85.5 \times× 0.34 = 29.07 g / L.$

=29.07 g / L.