Some fertilizer blends contain magnesium nitrate (MgNO3). Suppose that a chemist has 1.24 liters of a 2.13 M solution of magnesium nitrate. If the chemist dilutes the solution to 1.60 M, what is the volume of the new solution?

Question

Lana Cherry

Chemistry

10 February, 15:03

Some fertilizer blends contain magnesium nitrate (MgNO3). Suppose that a chemist has 1.24 liters of a 2.13 M solution of magnesium nitrate. If the chemist dilutes the solution to 1.60 M, what is the volume of the new solution?

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Answers (2)

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Shyla Larson · Answer 1

Initial concentration of magnesium nitrate M1 = 2.13 M

Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L

Final concentration of MgNO3, M2 = 1.60 M

Let the final volume of MgNO3 upon dilution be V2

Formula to use:

M1*V1 = M2*V2

V2 = M1*V1/M2

= 2.13 M * 1.24 L/1.60 M = 1.65 L

I believe I did it right ...

Carissa Ferrell · Answer 2

Initial concentration of magnesium nitrate M1 = 2.13 M

Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L

Final concentration of MgNO3, M2 = 1.60 M

Let the final volume of MgNO3 upon dilution be V2

Formula to use:

M1*V1 = M2*V2

V2 = M1*V1/M2

= 2.13 M * 1.24 L/1.60 M = 1.65 L

Thus, the final volume of magnesium nitrate solution upon dilution is 1.65 L