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13 January, 17:25

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

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  1. 13 January, 19:19
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    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
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