A gaseous sample containing a mixture of only nitrogen andcarbon monoxide, occup
ID: 686109 • Letter: A
Question
A gaseous sample containing a mixture of only nitrogen andcarbon monoxide, occupies a volume of 3.05L at 22°C.The partial pressure of N2 is0.130 atm and that of CO is 0.355 atm. A) What is the total pressure of the mixture? B) What is the mole fraction ofN2 in the mixture? C) How many moles of N2 are in the sample? D) What is the total mass of the sample? A gaseous sample containing a mixture of only nitrogen andcarbon monoxide, occupies a volume of 3.05L at 22°C.
The partial pressure of N2 is0.130 atm and that of CO is 0.355 atm. A) What is the total pressure of the mixture? B) What is the mole fraction ofN2 in the mixture? C) How many moles of N2 are in the sample? D) What is the total mass of the sample?
Explanation / Answer
A) Total pressure of a gaseous mixture is equals tothe sum of the partial pressures of each. so total pressure =pratial pressure of N2 + partial pressure of CO = 0.130atm + 0.355atm = 0.485atm B) we know , pV = nRT nN2 = (pN2 *V) /(RT) = (0.130atm*3.05L) / (0.0821Latm.mol^-1K^-1 * (22+273)K) =0.0164mol similarly, nCO= (pCO *V) /(RT) = (0.355atm*3.05L) / (0.0821Latm.mol^-1K^-1 * (22+273)K) = 0.0447mol Mole fraction of N2 = 0.0164mol / (0.0164mol +0.0447mol) = 0.2684 C) we know , pV = nRT nN2 = (pN2 *V) /(RT) = (0.130atm*3.05L) / (0.0821Latm.mol^-1K^-1 * (22+273)K) =0.0164mol nN2 = (pN2 *V) /(RT) = (0.130atm*3.05L) / (0.0821Latm.mol^-1K^-1 * (22+273)K) =0.0164mol D) mass = moles * molar mass so total mass of sample = 0.0164mol*(28g/mol) + 0.0477mol *(28g/mol) = 1.7108gRelated Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.