Gen Chem, Factors Affecting Solubility

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SUMMARY

The discussion focuses on calculating the mass of nitrogen dissolved in an 80.0L aquarium at room temperature, given a total pressure of 1.0 atm and a nitrogen mole fraction of 0.78. The solubility equation used is Sgas = KH x Pgas, with KH specified as 6.1 x 10-4 M/atm. The initial calculation yielded 1.37 g of N2, but the correct answer is 1.1 g N2, indicating an error in the application of the mole fraction and partial pressure concepts.

PREREQUISITES
  • Understanding of gas solubility principles
  • Familiarity with Henry's Law and its application
  • Knowledge of mole fraction calculations
  • Basic chemistry concepts related to pressure and volume
NEXT STEPS
  • Review Henry's Law and its implications for gas solubility
  • Study mole fraction calculations in detail
  • Learn about partial pressure and its role in gas mixtures
  • Practice problems involving solubility in different conditions
USEFUL FOR

Chemistry students, educators, and anyone involved in aquatic chemistry or gas solubility calculations will benefit from this discussion.

coffeecat91
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Homework Statement



Calculate the mass of nitrogen dissolved at room temperature in an 80.0L home aquarium. Assume a total pressure of 1.0 atm and a mole fraction for nitrogen of .78.

Homework Equations



Mole fraction= moles solute/moles solution

Sgas=KH x Pgas

KH = 6.1 x 10-4 M/atm

The Attempt at a Solution



Sgas = (6.1 x 10-4)(1 atm)
Sgas = 6.1 x 10-4
6.1 x 10-4/ 1 L x 80.0 L = 4.88 x 10-2 moles
(4.88 x 10-2 moles) x 28.02 g Ns/ 1 mol = 1.37 g N2

the correct answer is 1.1 g N2
Where did I go wrong and where does the mole fraction bit come into play? Thanks!
 
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coffeecat91 said:
Where did I go wrong and where does the mole fraction bit come into play?

Partial pressure.
 

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