You have 67.0 mL of a 0.400 M stock solution that must be diluted to 0.100 M. Assuming the volumes are additive, how much water should you add?
you want 1/4 the concentration, so you need 4 times the volume.
So, add 3*67.0mL of water
If 4.30 g of CuNO3 is dissolved in water to make a 0.560 M solution, what is the volume of the solution?
if 4.30 g of CuNO3 is dissolved in water to make a 0.560 M solution, what is the volume of the solution?
To determine how much water should be added to the 0.400 M stock solution in order to obtain a 0.100 M solution, we can use the formula for dilution:
C1V1 = C2V2
where C1 and V1 are the concentration and volume of the initial stock solution, and C2 and V2 are the concentration and volume of the final diluted solution.
We are given:
C1 = 0.400 M (concentration of the stock solution)
V1 = 67.0 mL (volume of the stock solution)
C2 = 0.100 M (desired concentration of the diluted solution)
V2 = ?
Plugging in these values into the formula, we have:
(0.400 M)(67.0 mL) = (0.100 M)(V2)
Solving for V2, we divide both sides of the equation by 0.100 M:
(0.400 M)(67.0 mL) / (0.100 M) = V2
Simplifying, we get:
V2 = (0.400 M)(67.0 mL) / (0.100 M)
V2 = 268.0 mL
Therefore, in order to dilute the 0.400 M stock solution to a concentration of 0.100 M, you should add 268.0 mL of water.