Concentration of potassium dichromate is equal to 0,02555 mol/l. Initial
volume is 250 ml. Till which volume (in ml) it should be diluted, as to receive
solution with desired titrimetric conversion factor of potassium dichromate
corresponding to Iron(II) (M=55,847 g/mol) to be equal to 3,839∙10-3
g/ml?Express numerical result with an accuracy of: X,X
To calculate the volume to which the solution should be diluted, we need to consider the desired titrimetric conversion factor and the concentration of potassium dichromate.
The molar mass of potassium dichromate is 294.18 g/mol (2 x 39.1 g/mol for potassium + 2 x 51.996 g/mol for chromium + 7 x 16 g/mol for oxygen).
Given that the conversion factor is 3.839×10^(-3) g/ml, we can calculate the molar mass of iron(II).
Molar mass of iron(II) = 3.839×10^(-3) g/ml / (0.02555 mol/l) = 0.15014 g/mol
Now, we can calculate the moles of iron(II) using its molar mass.
Number of moles of iron(II) = 0.15014 g/mol / 55.847 g/mol = 0.002688 mol
Since the desired conversion factor is the same for both potassium dichromate and iron(II), the moles of potassium dichromate should also be 0.002688 mol.
To find the volume to which the solution should be diluted, we can rearrange the equation for molarity:
Molarity (M) = Moles of solute / Volume of solution (in liters)
0.02555 mol/l = 0.002688 mol / V (in liters)
Solving for V:
V (in liters) = 0.002688 mol / 0.02555 mol/l = 0.1052 l
Finally, we convert the volume to milliliters:
V (in ml) = 0.1052 l * 1000 ml/l = 105.2 ml
Expressed with an accuracy of one decimal place, the volume to which the solution should be diluted is 105.2 ml.