Question

A stock solution of potassium nitrate, KNO3, has a concentration of 0.67 M. What volume in mL of dilute potassium nitrate (0.10M) can be formed with 73.04 mL of the concentrated
solution? Solve to 2 decimal places.​

Answer 1

Answer:

489.368 mL

Explanation:

given stock solution concentration, M1 = 0.67 M

stock solution volume used, V1 = 73.04 mL

given concentration of diluted solution, M2 = 0.10 M

we need to calculate volume of diluted solution = V2

applying law of dilution

M2 × V2 = M1 × V1

=> 0.10 M × V2 = 0.67 M × 73.04 mL

=> V2 = 489.368 mL

so volume of diluted solution is 489.368 mL

Thank you!