Question
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
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
Answers
GPT 3.5
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.
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.