Calculate the volume of 2.0 mol dm^-3 sulphuric acid,H2SO4 needed to prepare 2.5 dm^3 of 0.5 mol dm^-3 of the same acid solution.
Use:
C1V1=C2V2
Measure all concentrations in mol/dm^3, and volumes in dm^3
2.0*V = 0.5*2.5
V=1.25/2=0.625 dm^3
or, approximately,
0.63 dm^3 rounded to 2 significant figures.
To calculate the volume of the 2.0 mol dm^-3 sulphuric acid needed, we can use the equation:
(C1)(V1) = (C2)(V2)
where C1 and V1 are the concentration and volume of the initial solution, and C2 and V2 are the concentration and volume of the final solution.
Given:
C1 = 2.0 mol dm^-3
V1 = ?
C2 = 0.5 mol dm^-3
V2 = 2.5 dm^3
Substituting the values into the equation:
(2.0 mol dm^-3)(V1) = (0.5 mol dm^-3)(2.5 dm^3)
Simplifying:
V1 = (0.5 mol dm^-3)(2.5 dm^3) / (2.0 mol dm^-3)
V1 = 1.25 dm^3
Therefore, you would need 1.25 dm^3 of 2.0 mol dm^-3 sulphuric acid to prepare 2.5 dm^3 of 0.5 mol dm^-3 of the same acid solution.
To calculate the volume of 2.0 mol dm^-3 sulphuric acid (H2SO4) needed to prepare 2.5 dm^3 of 0.5 mol dm^-3 of the same acid solution, we can use the equation:
(C1)(V1) = (C2)(V2)
Where:
C1 = initial concentration of the acid solution (2.0 mol dm^-3)
V1 = initial volume of the acid solution (unknown)
C2 = final concentration of the acid solution (0.5 mol dm^-3)
V2 = final volume of the acid solution (2.5 dm^3)
Rearranging the equation to solve for V1, we get:
V1 = (C2)(V2) / C1
Substituting the given values into the equation:
V1 = (0.5 mol dm^-3)(2.5 dm^3) / (2.0 mol dm^-3)
V1 = 1.25 dm^3
Therefore, to prepare 2.5 dm^3 of 0.5 mol dm^-3 sulphuric acid solution, you would need to use 1.25 dm^3 of 2.0 mol dm^-3 sulphuric acid.