Answer to Question #192352 in General Chemistry for lara

Question

Use the following information: A container has 8360.0 torr of Nitrogen, 5.00 atm of Oxygen, 101.1 kPa of Argon, and 760.0 mm Hg of Carbon Dioxide.

What is the mole fraction of Carbon Dioxide?

A. 0.0556

B. 0.0670

C. 0.2567

D. 0.6676

E. 0.7572

Answer 1

$\frac {8360 torr × 1 atm} {760 torr}$ = 11 atm N2

5 atm O2

$\frac {101.1 kPa × 1 atm} {101.32}$ = 0.998 atm Ar

$\frac {760 mm Hg × 1 atm} {760 mm Hg}$ = 1 atm CO2

Total Pressure = 11 atm + 5 atm + 0.998 atm +1 atm = 17.998 atm

mole fraction = $\frac {5 atm} {17.998}$

Mole fraction of Carbon Dioxide= 0.278