How many milliliters of stock solution of 2.00 M KNO3 would you need to prepare 100.0 ml of .20 M KNO3 solution?
the molarity is being reduced to a tenth of the original stock solution
so the stock solution is one tenth of the total prepared solution
Could you explain a bit more? Im sorry im just so confued on this one
To determine the amount of stock solution needed to prepare the desired solution, we can use the formula:
(M₁)(V₁) = (M₂)(V₂),
where:
M₁ = initial concentration of the stock solution,
V₁ = volume of the stock solution to be measured,
M₂ = desired concentration of the final solution, and
V₂ = final volume of the solution to be prepared.
In this case, we have:
M₁ = 2.00 M (concentration of the stock solution),
V₁ = unknown,
M₂ = 0.20 M (desired concentration of the final solution), and
V₂ = 100.0 ml (final volume of the solution to be prepared).
Plugging in these values into the formula, we have:
(2.00 M)(V₁) = (0.20 M)(100.0 ml).
Now, we can solve for V₁:
V₁ = (0.20 M)(100.0 ml) / (2.00 M).
By performing the calculation, we find that V₁ is equal to 10.0 ml.
Therefore, you would need 10.0 milliliters of the 2.00 M KNO3 stock solution to prepare 100.0 ml of 0.20 M KNO3 solution.