Question

A 6.00 L tank at 7.39 C is filled with 16,1 g of chlorine pentafluoride gas and 5.21 g of dinitrogen monoxide gas. You can assume both gases behave as ideal gases under these conditions. Calculate the mole fraction and partial pressure of each gas, and the total pressure in the tank. Round each of your answers to 3 significant digits. chlorine pentafluoride tm X 5 ? mole fraction: partial pressure: mole fraction: partial pressure: dinitrogen monoxide atm Total pressure in tank: atm

Answer 1

Step 1: find mole fraction of both components

Molar mass of ClF5,

MM = 1*MM(Cl) + 5*MM(F)

= 1*35.45 + 5*19.0

= 130.45 g/mol

Molar mass of N2O,

MM = 2*MM(N) + 1*MM(O)

= 2*14.01 + 1*16.0

= 44.02 g/mol

n(ClF5) = mass of ClF5/molar mass of ClF5

= 16.1/130.45

= 0.1234

n(N2O) = mass of N2O/molar mass of N2O

= 5.21/44.02

= 0.1184

n(ClF5),n1 = 0.1234 mol

n(N2O),n2 = 0.1184 mol

Total number of mol = n1+n2

= 0.1234 + 0.1184

= 0.2418 mol

Mole fraction of each components are

X(ClF5) = n1/total mol

= 0.1234/0.2418

= 0.5105

X(N2O) = n2/total mol

= 0.1184/0.2418

= 0.4895

Step 2: find total pressure

Given:

V = 6.0 L

n = 0.2417742275062609 mol

T = 7.39 oC

= (7.39+273) K

= 280.39 K

use:

P * V = n*R*T

P * 6 L = 0.2418 mol* 0.08206 atm.L/mol.K * 280.39 K

P = 0.9272 atm

Step 3: find partial pressure of both components

p(ClF5) = X(ClF5) * P

= 0.5105 * 0.9272

= 0.4733 atm

p(N2O) = X(N2O) * P

= 0.4895 * 0.9272

= 0.4539 atm

Answer:

mole fraction of ClF5 = 0.510

p(ClF5) = 0.473 atm

mole fraction of N2O = 0.490

p(N2O) = 0.454 atm

p Total = 0.927 atm