What is the mass of K2SO4.Cr2(SO4)3.24H2O (Relative molecular mass = 894) required to make 1 dm3 of a 10.4 ppm Cr3+ solution (Cr-52)? A) 8.94 mg B) 8.94 g C) 17.88 mg D) 178.8 mg E) 89.4 mg

First, we need to understand what ppm (parts per million) means. In this case, 10.4 ppm Cr3+ means there are 10.4 mg of Cr3+ in 1 dm3 of solution.

Next, we need to calculate the proportion of Cr3+ in the compound K2SO4.Cr2(SO4)3.24H2O. The relative atomic mass of Cr is 52, and there are two Cr atoms in the compound, so the total mass of Cr in the compound is 2*52 = 104.

The proportion of Cr3+ in the compound is therefore 104/894 = 0.11633.

To make a 10.4 ppm solution, we need 10.4 mg of Cr3+. To find out how much of the compound we need to get this amount of Cr3+, we divide 10.4 by the proportion of Cr3+ in the compound: 10.4 / 0.11633 = 89.4 mg.

So, the answer is E) 89.4 mg.