A 200 mL K2SO4 solution with a concentration of 0.40 M is mixed with 400 mL of a K2SO4 solution containing 7% K2SO4 (m/v). The molar mass of K2SO4 is 174.3 g/mol. What is the final concentration of K2SO4 in the solution in molarity

The final concentration in molarity (M) is 0.4

Explanation:

K₂SO₄ (Potassium sulfate) - Molar mass = 174.3 g/m

First of all, we have 200 mL, 0.4 M

Molarity. volume = mol

(mol/L). L = mol → 0.4 mol/L. 0.2L = 0.08 moles of salt, in the beginning.

In second situation we have 7% m/v of salt, which means that in 100 mL of solution, we have 7 grams of solute.

7 g / 174.3 g/m = 0.040 moles in 100 mL

Let's think the rule of three:

In 100 mL we have 0.04 moles of salt

in 400 mL we have (400. 0.04) / 100 = 0.16 moles

1st situation: 0.08 moles of salt

2nd situation: 0.16 moles of salt

Total moles = 0.08 + 0.16 = 0.24 moles

Total volume = 200 mL + 400 mL = 600 mL

Molarity = mol/L

0.24 mol / 0.6L = 0.4M