2.00 L of 0.800 M NaNO3 must be prepareds from a solution which 1.50 M in concentration. How many mL of the 1.50 M are required ?
I am not sure weather " 2.00 L " is my V1 or V2 .
yor v2 cuz it must be prepared from 1.50
To find the volume of the 1.50 M solution required, we need to identify which volume (V1 or V2) represents the volume of the 1.50 M solution needed.
In this case, "2.00 L" refers to V2, which is the final desired volume (volume of the prepared solution).
V1 represents the initial volume or concentration of the solution available, which is 1.50 M in this case. So, the concentration of the available solution is 1.50 M (V1).
Therefore, V2 (final volume) = 2.00 L.
Now, we can use the formula for dilution to find the volume of the 1.50 M solution required:
M1V1 = M2V2
The initial molarity (M1) is 1.50 M, the initial volume (V1) is what we're trying to find, the final molarity (M2) is 0.800 M, and the final volume (V2) is 2.00 L.
1.50 M (V1) = 0.800 M (2.00 L)
Now, we can solve for V1:
V1 = (0.800 M) (2.00 L) / 1.50 M
V1 ≈ 1.0667 L or 1066.7 mL (rounded to four decimal places)
Therefore, approximately 1066.7 mL of the 1.50 M solution is required to prepare 2.00 L of 0.800 M NaNO3.