When a substance is dissolved in a solution, it’s rate of dissolution is jointly proportional to
(i) the amount of the substance still undissolved.
(ii) the difference between the concentration of the substance in a saturated solution and the actual concentration of the substance in the solution.
Now 6 grams of sulphur is placed in a solution of 100 cc of benzol which when saturated will hold 10 grams of sulphur. If 3 grams of sulphur is dissolved in the solution in 50 minutes, how many grams will be dissolved in the solution in 100 minutes?

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answer is 4.23.

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Detailed Solution

Let x gm of sulphur yet to be dissolved at time t.
$\frac{d x}{d t} = k x (0.1 - \frac{6 - x}{100}) = \frac{k x (4 + x)}{100}$
$\int_{6}^{x} \frac{d x}{x (4 + x)} = \int_{0}^{T} \frac{K d t}{100}$
$\log \frac{5 x}{3 (x + 4)} = \frac{K T}{25}$
$@ T = 50, x = 3$
To find $x @ T = 100.$
Solving $x = \frac{30}{17}$
So sulphur dissolved = $6 - \frac{30}{17} = \frac{72}{17} g m .$