Question

A stock solution is made by dissolving 66.05 g of (NH4)2SO4 in enough water to make 250 mL of solution. A 10.0 mL sample of this solution is then diluted to 50.0 mL. Given that the molar mass of (NH4)2SO4 is 132.1 g/mol, what is the concentration of the new solution? Use mc022-1.jpg and mc022-2.jpg. 0.400 M 1.60 M 5.00 M 10.0 M

Answer 1

first the stock solution is prepared

we need to calculate the molar concentration of (NH₄)₂SO₄

mass of (NH₄)₂SO₄ added - 66.05 g

number of moles of (NH₄)₂SO₄ - 66.05 g / 132.1 g/mol = 0.5 mol

volume of the solution - 250 mL

molarity of solution is - number of moles / volume of solution

molarity of (NH₄)₂SO₄ - 0.5 mol / 0.250 L = 2 mol/L

10.0 mL of the stock solution is taken and diluted upto 50.0 mL

the dilution formula is used

c1v1 = c2v2

where c1 - concentration and v1 - volume of stock solution

c2 - concentration and v2 is volume of the diluted solution

substituting the values

2 mol/L x 10.0 mL = C x 50.0 mL

C = 0.4 mol/L

concentration of new solution is 0.4 M

Answer 2

So we start with finding the molarity of the stock solution.
66.05g / 132.1g/mol = 0.5mol / .250L = 2M
Then it is diluted so you use M1V1 = M2V2 to solve for the new molarity.
2M x 10mL = M2 x 50mL
M2 = 0.4M
So your new molarity is 0.400M.