What volume (in mL) of 0.190 M sodium thiosulphate, Na2S2O3, solution is needed to prepare 250.0 mL of 0.0140 M sodium thiosulphate solution?
mL1 x M1 = mL2 x M2
mL1 x 0.190 = 250.0 x 0.0140
Solve for mL1 = ?
.2500 L * 0.0140 mol/L = ? mol
? mol / 0.190 mol/L = ? L
To calculate the volume of the 0.190 M sodium thiosulphate solution needed, we can use the equation:
C1V1 = C2V2
Where:
C1 = initial concentration of the sodium thiosulphate solution
V1 = initial volume of the sodium thiosulphate solution
C2 = final concentration of the sodium thiosulphate solution
V2 = final volume of the sodium thiosulphate solution
So, rearranging the equation, we have:
V1 = (C2 * V2) / C1
Substituting the values:
C1 = 0.190 M
C2 = 0.0140 M
V2 = 250.0 mL
V1 = (0.0140 M * 250.0 mL) / 0.190 M
V1 = 18.421 mL
Therefore, you would need approximately 18.421 mL of the 0.190 M sodium thiosulphate solution to prepare 250.0 mL of the 0.0140 M sodium thiosulphate solution.
To find the volume of 0.190 M sodium thiosulphate solution needed, we can use the formula:
(C1)(V1) = (C2)(V2)
Where:
C1 = concentration of the initial solution
V1 = volume of the initial solution
C2 = concentration of the final solution
V2 = volume of the final solution
We are given:
C1 = 0.190 M
V1 = unknown (what we need to find)
C2 = 0.0140 M
V2 = 250.0 mL
Plugging in these values into the formula, we have:
(0.190 M)(V1) = (0.0140 M)(250.0 mL)
To solve for V1, we can rearrange the equation:
V1 = [(0.0140 M)(250.0 mL)] / (0.190 M)
V1 ≈ 18.421 mL
Therefore, approximately 18.421 mL of the 0.190 M sodium thiosulphate solution is needed to prepare 250.0 mL of 0.0140 M sodium thiosulphate solution.