What is the concentration of N2 in a fresh water stream in equilibrium with air at 298 K and 1 atmosphere? Given the value of KH for N2 = 0.00060 mole/kgbar.(a) 0.0474 g/kg(b) 0.0005 g/kg(c) 1316.7 g/kg(d) 13.3 g/kgThe question was posed to me in exam.My question is taken from Solubility Solutions topic in chapter Solutions of Chemistry – Class 12

What is the concentration of N2 in a fresh water stream in equilibrium

1.	What is the concentration of N2 in a fresh water stream in equilibrium with air at 298 K and 1 atmosphere? Given the value of KH for N2 = 0.00060 mole/kgbar.(a) 0.0474 g/kg(b) 0.0005 g/kg(c) 1316.7 g/kg(d) 13.3 g/kgThe question was posed to me in exam.My question is taken from Solubility Solutions topic in chapter Solutions of Chemistry – Class 12
Answer» Right option is (d) 13.3 g/kg The BEST I can explain: Given, KH = 0.00060 mole/kgbar PN2 = mole fraction of N2 x Pair (from Dalton’s law) Air CONSISTS of 79 mole% N2 and 21 mole% O2. PN2 = 0.79 x 1 BAR= 0.79 bar Henry’s law –PN2x KH= solubility of N2 0.79 bar x 0.00060 (mole/kgbar) = solubility of N2 Solubility of N2 = 4.74 x 10^-4 moles of N2/kg water Converting moles of N2 to kg of N2 : Solubility of N2 = 4.74 x 10^-4 mole x 28 kg/mole = 0.0133 kg N2/kg water = 13.3 g N2/kg water.

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