a. How many milliliters of a stock solution of 6.0 M HNO3 would be required to prepare 110 mL of 0.500 M HNO3 solution?
b. IF you dilute 10.0 mL of the stock solution (referenced in part a) to a final volume of 0.250 L, what will be the concentration of the diluted solution?
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a. To find the amount of milliliters of the stock solution of 6.0 M HNO3 required to prepare 110 mL of 0.500 M HNO3 solution, we can use the formula:
M1V1 = M2V2
Where:
M1 = initial concentration of the stock solution
V1 = volume of the stock solution
M2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
Rearranging the formula to solve for V1:
V1 = (M2 * V2) / M1
Substituting the given values:
M1 = 6.0 M
V2 = 110 mL = 0.110 L
M2 = 0.500 M
V1 = (0.500 M * 0.110 L) / 6.0 M
V1 = 0.00917 L = 9.17 mL
Therefore, 9.17 mL of the stock solution of 6.0 M HNO3 would be required to prepare 110 mL of 0.500 M HNO3 solution.
b. To find the concentration of the diluted solution when diluting 10.0 mL of the stock solution to a final volume of 0.250 L, we can use the formula:
M1V1 = M2V2
Where:
M1 = initial concentration of the stock solution
V1 = volume of the stock solution
M2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
Substituting the given values:
M1 = 6.0 M
V1 = 10.0 mL = 0.010 L
V2 = 0.250 L
M2 = (M1 * V1) / V2
M2 = (6.0 M * 0.010 L) / 0.250 L
M2 = 0.240 M
Therefore, the concentration of the diluted solution will be 0.240 M.
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