Question

What volume of a 2.46 M

magnesium nitrate (Mg(NO3)2)
solution would be needed to make
275 mL of a 0.758 M solution by
dilution?
[ ? ] mL of 2.46 M Mg(NO3)2

Answer 1

84.73 mL of 2.46 M Mg(NO₃)₂

Step-by-step explanation:

To calculate the volume of 2.46 M magnesium nitrate [Mg(NO₃)₂] solution needed to make 275 mL of a 0.758 M solution by dilution, we can use the dilution formula.

Dilution formula

`\boxed{\sf C_1V_1=C_2V_2}`

where:

• C₁ = Initial concentration
• V₁ = Initial volume
• C₂ = Final concentration
• V₂ = Final volume

In this case, we are diluting the 2.46 M magnesium nitrate [Mg(NO₃)₂] solution to a concentration of 0.758 M, where the final volume is 275 mL, so:

• C₁ = 2.46 M
• C₂ = 0.758 M
• V₂ = 275 mL

Substitute these values into the formula and solve for V₁:

`\implies \sf 2.46\;M \cdot V_1=0.758\; M \cdot 275\;mL`

`\implies \sf V_1=(0.758\; M \cdot 275\; mL)/(2.46\;M)`

`\implies \sf V_1=84.735772...\;mL`

`\implies \sf V_1=84.73\;mL\;(2\;d.p.)`

Therefore, we need 84.73 mL of the 2.46 M magnesium nitrate [Mg(NO₃)₂] solution to make 275 mL of a 0.758 M solution by dilution.