How many grams of potassium di chromate (K2Cr2O7) are present in 1 lit of its N/10 solution in acid medium?
gram eq = NV = g/E = xg/M
The Attempt at a Solution
gram eq = NV = 0.1 * 1 = 0.1
g = gram eq * mol wt / valence factor (The valence factor I have taken as 14 because there are 7 oxygen atoms in potassium dichromate, which means in acidic solution there would be 14 H+ atoms)
0.1 * 294 / 14 = 2.1
How to proceed beyond this? The answer is one of the four:4.9, 49, 0.49, 3.9.