Answer to Question #349563 in General Chemistry for Siyabonga

Question

A 1.00 L flask contains 2.073 g O2 and 20.13 g N2 at 25.0°C. What is the total pressure? [1] 0.856 atm

[2] 1.09 atm

[3] 19.2 atm

[4] 22.2 atm

[5] None of the above 

Answer 1

V = 1 L

T = 25°C = 298 K

m(O₂) = 2.073 g

m(N₂) = 20.13 g

P = ?

n(O₂) = m / M = 2.073 / 32 = 0.065 mol

n(N₂) = m / M = 20.13 / 28 = 0.72 mol

n_total = n(O₂) + n(N₂) = 0.785 mol

$P =~ \cfrac {nRT}{V}=~ \cfrac {0.785*8.31*298}{1}=~1944~ kPa$

1 atm = 101.3 kPa

X = 1944 kPa

X = 19.2 atm

Answer: [3] 19.2 atm