How many milliliters of a solution of 8.8M CaBr2 are needed to prepare 175 mL of a 2.2M solution?

The product of molarity and volume of any substance at a fixed temperature is constant. Therefore comparing two portions of different molarity, we can say that, M₁V₁=M₂V₂

M represents the molarity, V represents volume of the solution.

M₁ in the question is 8.8M

M₂ is 2.2M

V₁=?

V₂ = 175

therefore,

8.8ₓV₁ = 2.2ₓ175ml

V₁ = (2.2ₓ175ml) / 8.8

=43.75 ml