Question

A solution with a total volume of 1000.0 mL contains 37.1 g Mg(NO3)2. If you remove 20.0 mL of this solution and then dilute this 20.0 mL sample with water until the new volume equals 500.0 mL, what is the concentration of Mg 2 ion in the 500.0 mL of solution?

Answer 1

Answer:

0.01023 M  is the concentration of Mg 2 ion in the 500.0 mL of solution.

Explanation:

Mass of magnesium nitrate = 37.1 g

Moles of magnesium nitrate =$\frac{37.1 g}{148 g/mol}=0.02559 mol$

Volume of the solution = 1000.0 mL = 1.0 L ( 1000 mL = 1 L)

Molarity of the solution = $\frac{\text{Moles of solute}}{\text{Volume of solution} (L)}$

$M=\frac{0.02559 mol}{1.0 L}=0.02559 M$

Molarity of the magnesium nitrate solution before dilution = $M_1=0.02559 M$

Volume of the magnesium nitrate solution before dilution = $V_1=20.0 mL$

Molarity of the magnesium nitrate solution after dilution = $M_2=?$

Volume of the magnesium nitrate solution after dilution = $V_2=500.0 mL$

$M_1V_1=M_2V_2$(dilution )

$M_2=\frac{M_1V_1}{V_2}$

$=\frac{0.02559 M\times 20.0 mL}{500.0 mL}=0.01023 M$

Molarity of the magnesium nitrate solution after dilution is 0.01023 M.

$Mg(NO_3)_2(aq)\rightarrow Mg^{2+}(aq)+2NO_3^{-}(aq)$

1 mole of magnesium nitrate gives 1 mole of magnesium ions.Then in 0.01023 M solution of magnesium nitrate;

$[Mg^{2+}]=1\times 0.01023 M=0.01023 M$

0.01023 M  is the concentration of Mg 2 ion in the 500.0 mL of solution.