Solving Potassium Dichromate Homework Problem

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To determine the grams of potassium dichromate (K2Cr2O7) in a 1 liter N/10 solution, the calculation starts with the formula gram eq = NV, yielding 0.1 equivalents. The valence factor is considered as 14 due to the acidic medium, leading to the equation g = gram eq * mol wt / valence factor. This results in 2.1 grams, but further calculations with the equivalent weight of 49 grams per equivalent suggest that the correct answer is 4.9 grams. The final conclusion confirms that 4.9 grams is indeed the correct amount of potassium dichromate in the solution.
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Homework Statement


How many grams of potassium di chromate (K2Cr2O7) are present in 1 lit of its N/10 solution in acid medium?

Homework Equations



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.
 
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OK, got it.

gram eq = no. of grams / Equ. wt

E = mol. wt/val factor = 294.2 / 6 = 49

g.e * E = 4.9.

Is this the answer?
 
Looks OK to me.
 

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