What volume of (dm\(^3\)) of water will be added to 10dm\(^3\) of 2.0 mol/dm\(^3\) HCL...

What volume of (dm\(^3\)) of water will be added to 10dm\(^3\) of 2.0 mol/dm\(^3\) HCL acid solution to give a final solution of 0.5 mol/dm\(^3\)?

To make a fixed amount of a dilute solution from a stock solution,

use the formula: C\(_1\)V\(_1\) = C\(_2\)V\(_2\)

where: V\(_1\) = Volume of stock solution needed to make the new solution.

C\(_1\) = Concentration of stock solution.

V\(_2\) = Final volume of new solution.

C\(_2\)= Final concentration of stock solution

2 * 10 = 0.5 * V2

20 = 0.5V2

\(\frac{20}{0.5}\) = V2

V\(_2\)= 40dm\(^3\)

What would be added = V\(_2\) - V\(_1\)

= 40 - 10 = 30dm\(^3\)

There is an explanation video available below.

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