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sergejj [24]
3 years ago
15

Some fertilizer blends contain magnesium nitrate (MgNO3). Suppose that a chemist has 1.24 liters of a 2.13 M solution of magnesi

um nitrate. If the chemist dilutes the solution to 1.60 M, what is the volume of the new solution?
Chemistry
2 answers:
tamaranim1 [39]3 years ago
7 0

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

Eduardwww [97]3 years ago
6 0

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 = <em>1.65 L</em>

<em>I believe I did it right...</em>

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