Part A Exercise 5.65 Enhanced with Feedback Calculate the mole fraction of N2 A

Question

Part A Exercise 5.65 Enhanced with Feedback Calculate the mole fraction of N2 A gas mixture contains 1.35 g N2 and 0.85 g O2 in a 1.61-L container at 18°C. Express your answer to three significant figures You want to reference ( pages 216-222) Section 5.6 while completing this problem XN2 Submit My Answers Give Up Part B Calculate the mole fraction of O2 Express your answer using two significant figures x2= 10.373 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Part C Calculate the partial pressure of N2 Express your answer using two significant figures atm

Explanation / Answer

At first you need to convert the given amount of N2 and O2 from grams to moles.

The molecular mass of N2 is 28.0134 g/mole.

The molecular mass of O2 is 31.9988g/mole.

so number of moles of N2 is = 1.35 g N2*(1 mole N2/28.0134 g N2) = 0.0482 mole of N2.

so number of moles of O2 is = 0.85g O2 *(1mole O2/31.9988g O2) = 0.0265 mole of O2.

therefore mole fraction O2 = X(O2) = moles of O2/(moles of N2 + moles of O2)

X(O2) = 0.0265mole / ( 0.0747mole) = 0.355 (solution of Part B)

similarly the mole fraction of N2 = X(N2) = moles of N2 /(moles of N2+moles of O2)

X(N2) = 0.0482 mole / (0.0747mole) = 0.645 (solution of part A)

The partial pressure of N2 is = number of moles of N2 *R * T / volume

here T = 18 degree C = 18+273.15 K = 291.15 K .

volume = V = 1.61 L

R = universal gas constant = 0.0821 L atm / K mole

therefore partial pressure of N2 = P(N2) = 0.0482 mole of N2 *(0.0821 L atm / K mole) * (291.15 K) / 1.61 L

P(N2 )= 0.72 atm ( solution of part c )

similarly by following the above method the Partial pressure of O2 was found out to be

P(O2) = 0.0265 mole of O2 * (0.0821 L atm / K mole) * (291.15 K) / 1.61 L

P(O2) = 0.39 atm (solution of Part D).

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