A sample of 1500 mL of 00500 M NH42S04 is diluted to 10000 m

A sample of 15.00 mL of 0.0500 M (NH42S04 is diluted to 100.00 mL in a volumetric flask. Then 6.00 mL of this solution is removed and diluted further to a total volume of 17.00 mL. What is the molarity of NHs in the final solution?

Solution

M (molarity of the solution) = 0.0500 M

V (volume of the solution) = 15.00 mL

Number of moles in the i sample (n): 15.00 mL x 0.0500 M = 0.75 mmol (MV)

Volume (diluted to) = 100 mL

Concentration of the solution = 0.075 mmol / 100 mL = 7.5x 10^-3 mol/L

Solution used (removed) = 6.00 mL = 6.00 x 10^-3 L

Amount of (NH₄)₂SO₄ present in 6.00 mL = 7.5x 10^-3 mol/L x 6.00 x 10^-3 L = 4.5x 10^-5 mol

Total volume of the solution in the second dilution = 17.00 mL = 17 x 10^-3 L

Molarity of the final solution = 4.5x 10^-5 mol /17 x 10^-3 L = 0.26x10^-2 M = 0.0026 M