how much water should be added to 8 mL of 6% saline solution to reduce the concentration to 4%?

Answer 1

If V₁ is the volume of the first solution and C₁ is the concentration of that first solution and if V₂ is the volume of the second solution and C₂ is the concentration of that second solution then:

V₁•C₁ + V₂•C₂ = V_TC_T where V_T is the total volume and C_T is the concentration of that total.

So: V₁ = 8 and C₁ = 0.06 and V₂ = ? but C₂ = 0.00 (water has no saline) and V_T = ? and C_T = 0.04:

V₁•C₁ + V₂•C₂ = V_TC_T

(8)•(0.06) + V₂•(0.00) = V_T•(0.04)

0.48 + 0 = 0.04V_T

V_T = 0.48/0/0.04 = 12

If you start with 8 mL and end with 12 mL then you added 4 mL of water