Question

how much water must be added when 125 mL of a 2.00 M solution of HCl is diluted to a final concentration of 0.400M?

Answer 1

To determine the amount of water to be added, we use the equation:

M1V1 = M2V2

where M1 and M2 are the concentrations and V1 and V2 are volumes. We calculate as follows:

2.00 (125) = 0.4(V2)
V2 = 625 mL 

Water to be added = 625 - 125 = 500 mL of water

Hope this answers the question.

Answer 2

Answer: The volume of water added will be 500 mL

Explanation:

To calculate the volume of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted solution

We are given:

$M_1=2.00M\\V_1=125mL\\M_2=0.400M\\V_2=?mL$

Putting values in above equation, we get:

$2.00\times 125=0.400\times V_2\\\\V_2=625mL$

Volume of water added = $V_2-V_1=(625-125)mL=500mL$

Hence, the volume of water added will be 500 mL