Question

When preparing 150.0 mL of a 0.250 M solution of Ca(OH)₂ from a 4.9 M stock solution how much of the stock solution (in mL) do you need to add? Record to the appropriate amount of significant figures.

Answer 1

The equation required for this type of problem is, V, 5, = V₂ S₂ V = volume of the Solution s- Strength of the Solution Vi = unknown s=4.9M V2 = 150.0 ml S2 = 0.25M. i. V, X4,9 M = 150 ml x 0.25 M or, Vi= 150X0.25 ml = 7.653 ml. i'. To prepeare 150 ml of 0.250 M Solution of CaCOH), , 7.653 ml of stock solution (4.9M) is needed.