Question

A particular experiment requires 350.0ml of 0.100 M HCL. A 2.20 M Stock solution is available. What volume of stock solution is needed to prepare the required HCL solution?

Answer 1

From a formula for concentration the number of moles of HCl present in 350.0 ml (or 0.350 L) is calculated:

C=n/V,

n=C×V,

n=0.100×0.350=0.035 (mol).

This number of moles is present in the next volume of the stock solution:

V=n/C,

V=0.035/2.20=0.016 (L) = 16 (ml).

Answer: 16 ml of the stock solution is needed.