Mercury(II) sulfide, HgS, is practically insoluble in pure
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?
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
You need to specify a volume... The smallest quantity per Liter is given = 3E-25 gram.
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.
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.
I don't either. With no grams listed, there is no way to know how much water is required to form a saturated solution.
You forgot this part:
Molar masses
(in g/mol)
Hg 200.6
S 32.07
To determine the smallest quantity of pure water that can be used to make a saturated solution of HgS, we need to consider the solubility of HgS and the desired concentration of the solution.
The solubility of HgS is given as no more than 3×10^−25 g/L at 25 degrees C. This means that at most 3×10^−25 grams of HgS can dissolve in 1 liter of water.
To calculate the smallest quantity of water needed to make a saturated solution, we need to determine the amount of HgS we want to dissolve. Let's assume we want to dissolve 1 gram of HgS.
To convert 1 gram to moles, we divide by the molar mass of HgS. The molar mass of mercury (II) sulfide (HgS) is approximately 200.59 g/mol.
Number of moles of HgS = (1 g) / (200.59 g/mol) = 0.0049889 mol
Now, we can use the solubility to calculate the volume of water needed. Since the solubility is given per liter, we need to convert the moles of HgS to liters.
Volume of water = (0.0049889 mol) / (3×10^−25 g/L) = 1.66296×10^22 L
Therefore, the smallest quantity of water needed to make a saturated solution of HgS with 1 gram of HgS is approximately 1.66296×10^22 liters.