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