To calculate the concentration of the solution, we first need to convert the mass of HgI2 to moles using the molar mass of HgI2.
Molar mass of HgI2:
Hg: 200.59 g/mol
I: 126.9 g/mol x 2 = 253.8 g/mol
Total molar mass = 200.59 + 253.8 = 454.39 g/mol
Now, we can calculate the number of moles of HgI2 in 0.011 g:
moles = mass / molar mass
moles = 0.011 g / 454.39 g/mol ≈ 2.420 x 10^-5 mol
Next, we need to calculate the concentration in mol/L:
Volume = 250. mL = 0.250 L
Concentration = moles / volume
Concentration = 2.420 x 10^-5 mol / 0.250 L ≈ 9.68 x 10^-5 mol/L
Therefore, the concentration of the chemist's mercury(II) iodide solution is approximately 9.68 x 10^-5 mol/L.
A chemist prepares a solution of mercuryII) iodide (HgI2) by measuring out 0.011 g of mercuryIl) iodide into a 250. mi volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mol/L of the chemist's mercury(II) iodide solution. Be sure your answer has the correct number of significant digits
1 answer