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 data:

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.

idk what is it? is this a question that you know the answer or u dont