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. 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?
Answer:
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