Question

WILL GIVE 50 POINTS PLEASE HELP!!!!!

Answer 1

Answer:

b) The volume of stock solution used is less than 3 liters.

Explanation:

Given:

Concentration of stock solution of NaCl, M1 = 10 M

Concentration of diluted solution of NaCl, M2 = 1 M

Volume of diluted NaCl solution, V2 = 3 L

Let the volume of stock NaCl solution = V1

Formula:

Use the dilution relation:

$M1V1 = M2V2$

Based on the above equation:

$V1 = \frac{M2V2}{M1} = \frac{1M*3L}{10M} = 0.3 L$

Therefore, the volume of stock solution required is 0.3 L which is less than 3 L