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?

Question

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?​

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Nho 2 weeks 2021-09-05T14:18:20+00:00 1 Answers 0 views 0

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    2021-09-05T14:19:32+00:00

    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

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