Solving Potassium Dichromate Homework Problem

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SUMMARY

The discussion focuses on calculating the grams of potassium dichromate (K2Cr2O7) in a 1-liter N/10 solution in an acidic medium. The solution utilizes the formula for gram equivalent weight, where the valence factor is determined to be 14 due to the presence of 7 oxygen atoms contributing to H+ ions in the acidic solution. The final calculation yields 4.9 grams of K2Cr2O7, confirming that this is the correct answer among the provided options.

PREREQUISITES
  • Understanding of normality and molarity in solutions
  • Familiarity with the concept of gram equivalent weight
  • Knowledge of potassium dichromate's molecular weight (294.2 g/mol)
  • Basic principles of acid-base chemistry
NEXT STEPS
  • Study the calculation of gram equivalent weight in various chemical reactions
  • Learn about the role of valence factors in acid-base titrations
  • Explore the applications of potassium dichromate in analytical chemistry
  • Review the principles of normality and its significance in solution preparation
USEFUL FOR

Chemistry students, educators, and professionals involved in analytical chemistry or solution preparation will benefit from this discussion.

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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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