Dilutions
6. If 30.0 mL of 12.0 M HCl stock solution are diluted to a volume of 500. mL, what is the molarity of the dilute solution?
7. If 27.5mL of 16.0 M nitric acid stock solution is added to water to make a 327.5mL solution, what is the molarity of the diluted solution? 1.34 M HNO³
8. If 50.0 mL of a stock sulfuric acid solution whose molarity is 15.0 M is diluted until the molarity of the new solution is 2.50 M, what is the volume of the new solution?
6. If 30.0 mL of 12.0 M HCl stock solution are diluted to a volume of 500. mL, what is the molarity of the dilute solution?
7. If 27.5mL of 16.0 M nitric acid stock solution is added to water to make a 327.5mL solution, what is the molarity of the diluted solution? 1.34 M HNO³
8. If 50.0 mL of a stock sulfuric acid solution whose molarity is 15.0 M is diluted until the molarity of the new solution is 2.50 M, what is the volume of the new solution?
Answers & Comments
Answer: PA BRAINLIEST
6. 0.72 M
7. 1.34 M
8. 300.0 mL
Explanation:
6. To calculate the molarity of the dilute solution, we can use the formula:
M1V1 = M2V2
Where M1 is the molarity of the stock solution, V1 is the volume of the stock solution used, M2 is the molarity of the dilute solution, and V2 is the final volume of the dilute solution. Plugging in the values we have:
M1 = 12.0 M
V1 = 30.0 mL = 0.030 L
V2 = 500. mL = 0.500 L
M2 = ?
12.0 M x 0.030 L = M2 x 0.500 L
M2 = (12.0 M x 0.030 L) / 0.500 L
M2 = 0.72 M
Therefore, the molarity of the dilute solution is 0.72 M.
7. Using the same formula as in problem 6:
M1 = 16.0 M
V1 = 27.5 mL = 0.0275 L
V2 = 327.5 mL = 0.3275 L
M2 = ?
16.0 M x 0.0275 L = M2 x 0.3275 L
M2 = (16.0 M x 0.0275 L) / 0.3275 L
M2 = 1.34 M
Therefore, the molarity of the diluted solution is 1.34 M.
8. Again, using the same formula:
M1 = 15.0 M
V1 = 50.0 mL = 0.050 L
M2 = 2.50 M
V2 = ?
15.0 M x 0.050 L = 2.50 M x V2
V2 = (15.0 M x 0.050 L) / 2.50 M
V2 = 0.30 L = 300.0 mL
Therefore, the volume of the new solution is 300.0 mL.