How many liters of 8.0M #HCl# are needed to prepare 1.50 L of 2.5M #HCl#?

1 Answer
Oct 10, 2016

Answer:

0.47 L of 8.0 M HCl are needed to prepare 1.50 L of 2.5 M HCl.

Explanation:

For dilution problems: #M_1V_1 = M_2V_2#, where #M_1# and #V_1# are the concentration and volume of the original solution and #M_2# and #V_2# are the concentration and volume of the diluted solution.

We are calculating #V_1#.

First rearrange the dilution equation:
#V_1 = (M_2V_2)/M_1#

Plug in the values:
#V_1 = ((2.5 M)(1.50 L))/(8.0 M)#

Calculate:
#V_1 = 0.47 L#