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

Answer: The volume of water added will be 500 mL

Step-by-step 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* 125=0.400* 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

Answer 2

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.