How many milliliters of a stock solution of 6.00 M HNO3 would you have to use to prepare 0.190 L of 0.490 M HNO3?
I used the equation M1V1=M2V2
6M*V1 = 0.190L*0.490 M
lead me to this V1=0.190L*0.490M/6.00 M
=0.0931 L per M/6.00M
=0.0155166L
0.0155166L/1000mL= 0.155166L
0.0155L * 1000mL/L = 15.5mL
Well, it looks like you did all the calculations correctly. So, you would need approximately 0.155 mL (or 0.155166 L) of the 6.00 M stock solution to prepare 0.190 L of 0.490 M HNO3. That's just a "drop" in the bucket!
To prepare 0.190 L of a 0.490 M HNO3 solution using a stock solution of 6.00 M HNO3, you would need to use 0.155166 L (or 155.166 mL) of the stock solution.
To calculate the number of milliliters (mL) of a stock solution of 6.00 M HNO3 needed to prepare 0.190 L of 0.490 M HNO3, you correctly used the equation M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
Plugging in the given values, the equation becomes:
6.00 M * V1 = 0.190 L * 0.490 M
First, you simplified the equation to:
V1 = (0.190 L * 0.490 M) / 6.00 M
Then, you simplified further to:
V1 = 0.0155166 L/M / 6.00 M
Finally, you converted liters (L) to milliliters (mL) by dividing by 1000:
V1 = 0.0155166 L/M / 6.00 M = 0.0155166 L/M * 1000 mL/L = 0.0155166 * 1000 mL = 15.5166 mL
Therefore, you would need approximately 15.5166 mL of the 6.00 M HNO3 stock solution to prepare 0.190 L of 0.490 M HNO3.