120 g of a 25% solution of magnesium sulfate was placed in a desiccator, where in a separate bowl, as a desiccant

120 g of a 25% solution of magnesium sulfate was placed in a desiccator, where in a separate bowl, as a desiccant, there was 100 g of 96% sulfuric acid. After some time, crystals of MgSO4 * 7H2O precipitated in the solution, the mass of which turned out to be equal to 24.6 g. Determine the mass fraction of sulfuric acid by the end of the experiment if the solubility of magnesium sulfate at room temperature is 35 g per 100 g of water

The solution of the problem:

35 / (35 + 100) = 0.2593 (mass fraction of MgSO4 in a solution saturated at room temperature);

120 * 0.25 = 30 (gram-weight of MgSO4 in the original solution);
120 * (24.6 / 246) = 12 (gram mass of MgSO4 in MgSO4 * 7H2O)

Let x gram of water from MgSO4 solution be bound by 96% H2SO4 solution. Then

30 – 12 / (120 – x – 24.6) = 0.2593;

x = 25.982 (gram);

100 * 0.96 / (100 + 25.982) = 0.762 = 76.2% (mass fraction of H2SO4)

Answer: 76.2%.