Question

If a chemist wishes to dilute a 1.000 × 10^3 mg/L stock solution to prepare 5.000 × 10^2 mL of a working standard that has a concentration of 7.000 mg/L, what volume of the 1.000 × 10^3 mg/L standard solution is needed?

Answer 1

Answer: 0.0035 L

Step-by-step explanation:

According to the dilution law,

`C_1V_1=C_2V_2`

where,

`C_1` = molarity of stock solution =
`1.000* 10^3mg/L`

`V_1` = volume of stock solution = ?

`C_2` = molarity of diluted solution =
`7.000mg/L`

`V_2` = volume of diluted solution =
`5.000* 10^2ml` = 0.5 L (1L=1000 ml)

Putting in the values we get:

`1.000* 10^3mg/L* V_1=7.000mg/L* 0.5L`

`V_1=0.0035L`

Therefore, volume of the
`1.000* 10^3mg/L` standard solution needed is 0.0035 L