How many milliliters of 16.0 M HNO3 stock solution wouuld be required to prepare 100.0mL of 3.00 M HNO3?
Use the formula M1V1=M2V2
So substitute and solve. What's the problem?
The solution that I got is 0.02 mL HNO3
To find the volume of the 16.0 M HNO3 stock solution required to prepare 100.0 mL of 3.00 M HNO3, we can use the formula M1V1 = M2V2, where M1 and V1 are the molarity and volume of the stock solution, and M2 and V2 are the desired molarity and volume of the final solution.
Let's assign the values to the variables:
M1 = 16.0 M (molarity of the stock solution)
V1 = volume of the stock solution (to be determined)
M2 = 3.00 M (desired molarity of the final solution)
V2 = 100.0 mL (desired volume of the final solution)
Now we can plug in the values into the formula and solve for V1:
M1V1 = M2V2
16.0 M * V1 = 3.00 M * 100.0 mL
Now, let's solve for V1:
V1 = (3.00 M * 100.0 mL) / 16.0 M
V1 = 300.0 mL / 16.0
Therefore, to prepare 100.0 mL of 3.00 M HNO3, we would need approximately 18.75 mL of the 16.0 M HNO3 stock solution.