what volume of a 0.5M solution of sulphuric acid is required to prepare 2 litres of a 0.4M solution of the acid
mL1 x M1 = mL2 x M2
mL1 x 0.5M = 2000 mL x 0.4
Solve for mL1.
To find out the volume of the 0.5M sulphuric acid solution needed to prepare 2 liters of a 0.4M solution, we can use the equation:
M1 × V1 = M2 × V2
Where:
M1 is the initial concentration (0.5M)
V1 is the initial volume (unknown)
M2 is the final concentration (0.4M)
V2 is the final volume (2L)
Rearranging the equation to solve for V1, we get:
V1 = (M2 × V2) / M1
Now, let's substitute the given values:
V1 = (0.4M × 2L) / 0.5M
Calculating this expression, we find:
V1 = 0.8L
Therefore, you will need 0.8 liters (or 800 milliliters) of the 0.5M sulphuric acid solution to prepare 2 liters of the 0.4M sulphuric acid solution.