6.67 mL of 18.0 molar sulfuric acid solution was dissolved in enough water to make 500 mL of solution. The molarity of the diluted solution is
To find the molarity of the diluted solution, we can use the equation:
M1V1 = M2V2
where:
M1 = initial molarity
V1 = initial volume
M2 = final molarity
V2 = final volume
In this case:
M1 = 18.0 M (initial molarity of sulfuric acid)
V1 = 6.67 mL (initial volume of sulfuric acid)
V2 = 500 mL (final volume of diluted solution)
Now, let's calculate the final molarity (M2):
M1V1 = M2V2
(18.0 M)(6.67 mL) = M2(500 mL)
120.06 mL·M = M2(500 mL)
M2 = (18.0 M)(6.67 mL) / (500 mL)
M2 ≈ 0.24 M
Therefore, the molarity of the diluted solution is approximately 0.24 M.
To find the molarity of the diluted solution, we need to use the formula:
M1V1 = M2V2
Where:
M1 = Molarity of the concentrated solution
V1 = Volume of the concentrated solution
M2 = Molarity of the diluted solution
V2 = Volume of the diluted solution
In this case, we have:
M1 = 18.0 M (given)
V1 = 6.67 mL = 0.00667 L (since 1 mL = 0.001 L)
V2 = 500 mL = 0.5 L (since 1 mL = 0.001 L)
Now we can substitute these values into the formula:
18.0 M × 0.00667 L = M2 × 0.5 L
To solve for M2 (molarity of the diluted solution), we divide both sides of the equation by 0.5 L:
(18.0 M × 0.00667 L) ÷ 0.5 L = M2
Calculating this gives us:
0.12006 M = M2
So, the molarity of the diluted solution is approximately 0.120 M.