Question

O len U. L < 19 of 32 > Review | Constants Periodic Table You may want to reference (Pages 328-330) Section 8.5 while completing this problem. Part A How would you prepare 2 L of 1 M MgCl, from a 3 M MgCl, stock solution? Add 330 mL of the 3 M MgCl, stock solution to enough distilled water for a total volume of 2 L of solution Add 1500 mL of the 3 M MgCl, stock solution to enough distilled water for a total volume of 2 L of solution. Add 670 mL of the 3 M MgCl, stock solution to enough distilled water for a total volume of 2 L of solution. Submit Request Answer Next > Provide Feedback

Answer 1

use dilution formula

M1*V1 = M2*V2

1---> is for stock solution

2---> is for diluted solution

Given:

M1 = 3 M

M2 = 1 M

V2 = 2 L

use:

M1*V1 = M2*V2

V1 = (M2 * V2) / M1

V1 = (1*2)/3

V1 = 0.67 L

V1 = 670 mL

So, we need 670 mL of original solution

Answer: option 3