Question

is 200 ml of 0.60 m mgcl2 is added to 400 ml of distilled water, what is the concentration of mg^2 in the resulting solution? (assume volumes are additive)

Answer 1

Concentration in resulting solution are 0.20 M Mg ion and 0.40 M Cl ion.

As volumes are additive, therefore, the total volume of the solution after adding 400 mL of distilled water is (200+400)mLor600mL. Hence, according to the law of dilution, we can write:

C1V1=C2V2

C1andC2 are the concentrations of initial and final MgCl2 solutions, respectively.

V1andV2 are the volumes of initial and final MgCl2 solutions, respectively.

Given:

V1=200mLC1=0.60MV2=600mL

Substituting the above values, we get:

0.60M×200mL=C2×600mL⇒C2=0.20M

1 mol of MgCl2 completely dissociates in aqueous solution to produce 1 mol of Mg2+(aq) and 2 mol of Cl−(aq) ions.

So, 0.20 M of MgCl2(aq) completely dissociates in aqueous solution to produce 0.20 M mol of Mg2+(aq) and 0.40 M mol of Cl−(aq) ions.

So, the concentrations of Mg2+(aq) and Cl−(aq) ions are 0.20 M and 0.40 M, respectively.

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