Mercury(II) sulfide, HgS, is practically insoluble in pure

Question

water. Its solubility at 25 degrees C is probably no more than
3×10^−25 g/L. Of the following quantities of pure water,
which is the smallest quantity that could be used to
make a saturated solution of HgS?

Answer 1

You need to specify a volume... The smallest quantity per Liter is given = 3E-25 gram.

Answer 2

You didn't list any quantities of pure water. From the problem 1 L would be a saturated solution but 1 L may not be listed as an answer and it may not be the smallest quantity.

Answer 3

This was a multiple chocie question, so the possible answers are:
A) 20,000 L
B) 1000 L
C) 10,000 L
D) 2000 L
E) 200 L

D, 2000 L, is apparently the correct answer, but I don't know how to come to that conclusion.

Answer 4

I don't either. With no grams listed, there is no way to know how much water is required to form a saturated solution.

Answer 5

You forgot this part:
Molar masses
(in g/mol)
Hg 200.6
S 32.07

Answer 6

This puzzled me for a while.

But find the mass of 1 mole of HgS [232.67g/mol]

Mass of 1 HgS unit is 232.67/6.023 x 10^23 = 3.863 x 10^-22 g
[divide mass of 1 mole by Avogadro number]

If solubility is 3 x 10^-25 g/L then solubility = mass/volume

so 3 x 10^-25 = 3.863 x 10^-22 / V

V = 1287L

So you need at least this volume so the smallest volume on the list that will make the saturate solution is 2000L