A stock solution of HNO3 is prepared and found to contain 11.1 M of HNO3. If 25.0 mL of the stock solution is diluted to a final volume of 0.500 L, the concentration of the diluted solution is ________ M.
0.555
To calculate the concentration of the diluted solution, we can use the formula for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
Given:
C1 = 11.1 M
V1 = 25.0 mL = 0.025 L (since 1 mL = 0.001 L)
V2 = 0.500 L
Plugging in the values:
(11.1 M)(0.025 L) = C2(0.500 L)
0.2775 M = C2
Therefore, the concentration of the diluted solution is 0.2775 M.
To find the concentration of the diluted solution, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the stock solution
V1 = initial volume of the stock solution
C2 = final concentration of the diluted solution
V2 = final volume of the diluted solution
Given information:
C1 = 11.1 M
V1 = 25.0 mL = 0.025 L
V2 = 0.500 L
Now we can substitute the values into the formula and solve for C2:
(11.1 M)(0.025 L) = C2(0.500 L)
To solve for C2, divide both sides by 0.500 L:
C2 = (11.1 M)(0.025 L) / 0.500 L
C2 = 0.555 M
Therefore, the concentration of the diluted solution is 0.555 M.
11.1 M x (25.0 mL/500 mL) = ?
OR use the dilution formula.
mL1 x M1 = mL2 x M2
25.0 x 11.1 = 500 mL x M2.
Solve for M2.