What volume of a 3.00M NaCl stock solution would you use to make 0.300 L of a 1.25M NaCl solution?

0.125 L

Explanation:

Given dа ta:

Initial volume = ?

Initial molarity of NaCl = 3.00 M

Final volume = 0.300 L

Final molarity = 1.25 M

Solution:

Formula:

M₁V₁ = M₂V₂

M₁ = Initial molarity

V₁ = Initial volume

M₂ = Final molarity

V₂ = Final volume

Now we will put the values in formula:

3.00 M * V₁ = 1.25 M * 0.300 L

3.00 M * V₁ = 0.375 M. L

V₁ = 0.375 M. L/3.00 M

V₁ = 0.125 L

0.125 L of 3.00M NaCl stock solution would be use to make 0.300 L of a 1.25M NaCl solution.