3. If you wish to make 500.0 mL of 0.150 M HCl solution from 6.00 M HCl stock solution, what volume of the stock would you need? 4) if 50.00 mL of 0.300 M AgNO3 are diluted to 250.0 mL, what is the concentration of the new solution?​

3. 12.5 mL

4. 0.06 M

Explanation:

3. Determination of the volume of stock solution.

Volume of diluted solution (V₂) = 500 mL

Molarity of diluted solution (M₂) = 0.150 M

Molarity of stock solution (M₁) = 6 M

Volume of stock solution needed (V₁) =?

M₁V₁ = M₂V₂

6 × V₁ = 0.150 × 500

6 × V₁ = 75

Divide both side by 6

V₁ = 75 / 6

V₁ = 12.5 mL

Thus, the volume of the stock solution needed is 12.5 mL

4. Determination of the molarity of the diluted solution.

Volume of stock solution (V₁) = 50 mL

Molarity of stock solution (M₁) = 0.3 M

Volume of diluted solution (V₂) = 250 mL

Molarity of diluted solution (M₂) =?

M₁V₁ = M₂V₂

0.3 × 50 = M₂ × 250

15 = M₂ × 250

Divide both side by 250

M₂ = 15 / 250

M₂ = 0.06 M

Thus, the concentration of the diluted solution is 0.06 M

93.2 ml + 3920 ml = 4013 ml total  

(93.2 ml) * (2.03 m) / (4013 ml) = 0.0471 m

the volume in in l of the diluted solution would be 0.0471 m.

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