**The answer is 0.0059 L, I would just like to know the process of how to get th
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Question
**The answer is 0.0059 L, I would just like to know the process of how to get that answer. Thanks!Use Henry's law and the solubilities given below to calculate the total volume of nitrogen and oxygen gas that should bubble out of 1.4 L of water upon warming from 25*C to 50*C . Assume that the water is initially saturated with nitrogen and oxygen gas at 25*C and a total pressure of 1.0 atm . Assume that the gas bubbles out at a temperature of 50*C . The solubility of oxygen gas at 50*C is 27.8 mg/L at an oxygen pressure of 1.00 atm. The solubility of nitrogen gas at 50*C is 14.6 mg/L at a nitrogen pressure of 1.00 atm. Assume that the air above the water contains an oxygen partial pressure of 0.21 atm and a nitrogen partial pressure of 0.78 atm.
Express your answer using two significant figures.
**The answer is 0.0059 L, I would just like to know the process of how to get that answer. Thanks!
Use Henry's law and the solubilities given below to calculate the total volume of nitrogen and oxygen gas that should bubble out of 1.4 L of water upon warming from 25*C to 50*C . Assume that the water is initially saturated with nitrogen and oxygen gas at 25*C and a total pressure of 1.0 atm . Assume that the gas bubbles out at a temperature of 50*C . The solubility of oxygen gas at 50*C is 27.8 mg/L at an oxygen pressure of 1.00 atm. The solubility of nitrogen gas at 50*C is 14.6 mg/L at a nitrogen pressure of 1.00 atm. Assume that the air above the water contains an oxygen partial pressure of 0.21 atm and a nitrogen partial pressure of 0.78 atm.
Express your answer using two significant figures.
Explanation / Answer
At 25 C and total pressure of 1.0 atm, Partial pressure of Oxygen, P O2 = 0.21 atm Partial pressure of nitrogen, P N2 = 0.78 atm . Given solubility at 50 C and 1 atm Solubility of oxygen = 27.8 mg/ L For nitrogen = 14.6 mg/ L . According to henry's law, Henry's law constant, k = solubility/ pressure . So k for oxygen = 27.8 mg/L / 1 atm = 27.8 mg/ (L.atm) For nitrogen = 14.6 mg/ L/ 1 atm =14.6 mg / (L.atm) At 25 C, solubility = k. P solubility of oxygen = 27.8 mg/ L.atm * 0.21 atm = 5.84 mg / L Similarlly solubility of nitrogen = 14.6 mg/ (L.atm) * 0.78atm =11.39 mg/ L We are given 1.9 L of water At 25 C, solubility = 5.84 mg / L mass of oxygen present in 1.9 L of water = 5.84 mg / L *1.9 L =11.1 mg At 50 C, solubility = 27.8 mg/ L Mass of oxygen = 27.8 mg/ L * 1.9 L =52.8 mg . Hence the difference in mass of oxgyen at the two temperatures= 52.8 mg - 11.1 mg = 41.7 mg This is the amount of oxygen which would be given out. . molar mass = 32 g/mol Pressure = 1 atm Temperature = 25 C = 298 K mass = 41.7 mg = 0.0417 g R = 0.0821 L-atm/ (mol.K) Volume, V = ( mass/ molar mass) * R * T / P Substitute and solve to get volume of oxygen. Similarly for nitrogen; At 25 C, solubility = 11.39 mg/ L Mass of nitrogen in 1.9 L = 11.39 mg / L * 1.9 L = 21.64mg . At 50 C solubility = 14.6 mg/ L Mass of nitrogen = 14.6 mg / L * 1.9 L = 27.74 mg . Mass of nitrogen released = 27.74 - 21.64 = 6.1 mg . molar mass = 28 g/mol Pressure = 1 atm Temperature = 25 C = 298 K mass = 6.1 mg = 0.0061 g R = 0.0821 L-atm/ (mol.K) Volume, V = ( mass/ molar mass) * R * T / P Substitute and solve to get volume of nitrogen. . Add both the volumes to get total volume of nitrogen andoxygen in Liters
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