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

Explanation:

According to the dilution law,

$C_1V_1=C_2V_2$

where,

$C_1$ = molarity of stock solution = $1.000\times 10^3mg/L$

$V_1$ = volume of stock solution = ?

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

$V_2$ = volume of diluted solution = $5.000\times 10^2ml$ = 0.5 L   (1L=1000 ml)

Putting in the values we get:

$1.000\times 10^3mg/L\times V_1=7.000mg/L\times 0.5L$

$V_1=0.0035L$

Therefore, volume of the $1.000\times 10^3mg/L$ standard solution needed is 0.0035 L