When 100ml of Hg(NO3)2 (aq) fully reacted with Na2S (s), 1g of HgS was producted. In this situation, how much is the molarity of Hg(NO3)??
Well I mean, the molarity of Hg(NO3)2. Not Hg(NO3)
1 g of HgS is how many mols of HgS
HgS is 233 g/mol (Hg = 201+ S= 32)
so we have 1/233 mol of HgS
That required 1/233 mol of Hg(NO3)2
(1/233)mol / 0.1 liter = 10/233 mol/liter = .043 M
Thanks a lot
You are welcome.
To determine the molarity of Hg(NO3)2, we need to use the given information.
From the balanced chemical equation:
Hg(NO3)2 + Na2S -> HgS + 2NaNO3
We can see that 1 mole of Hg(NO3)2 reacts with 1 mole of HgS.
Given that 1 gram of HgS is produced, we need to convert this mass into moles.
The molar mass of HgS is calculated by adding the atomic masses of mercury (Hg) and sulfur (S):
Atomic mass of Hg = 200.59 g/mol
Atomic mass of S = 32.06 g/mol
Molar mass of HgS = 200.59 g/mol + 32.06 g/mol = 232.65 g/mol
To convert the mass of HgS into moles, we divide the given mass by the molar mass:
Moles of HgS = 1 g / 232.65 g/mol ≈ 0.0043 mol
Since 1 mole of Hg(NO3)2 reacts with 1 mole of HgS, the number of moles of Hg(NO3)2 is also 0.0043 mol.
Now, we need to determine the molarity of Hg(NO3)2. Molarity is calculated by dividing the number of moles of solute (Hg(NO3)2 in this case) by the volume of the solution in liters.
Given that the volume of the solution is 100 mL (or 0.1 L):
Molarity of Hg(NO3)2 = Moles of Hg(NO3)2 / Volume of solution in liters
= 0.0043 mol / 0.1 L
= 0.043 M
Therefore, the molarity of Hg(NO3)2 is 0.043 M.