Question

A

10.00L

tank at

25.8°C

is filled with

7.32g

of chlorine pentafluoride gas and

7.82g

of carbon monoxide gas. You can assume both gases behave as ideal gases under these conditions.

Calculate the mole fraction of each gas. Be sure each of your answer entries has the correct number of significant digits.

Answer 1

Molar mass of ClF5 = 1*MM(Cl) + 5*MM(F)

= 1*35.45 + 5*19.0

= 130.45 g/mol

Molar mass of CO = 1*MM(C) + 1*MM(O)

= 1*12.01 + 1*16.0

= 28.01 g/mol

n(ClF5) = mass/molar mass

= 7.32/130.45

= 0.0561

n(CO) = mass/molar mass

= 7.82/28.01

= 0.2792

n(ClF5),n1 = 0.0561 mol

n(CO),n2 = 0.2792 mol

Total number of mol = n1+n2

= 0.0561 + 0.2792

= 0.3353 mol

Mole fraction of each components are

X(ClF5) = n1/total mol

= 0.0561/0.3353

= 0.1674

X(CO) = n2/total mol

= 0.2792/0.3353

= 0.8326

mole fraction of ClF5 = 0.167

mole fraction of CO = 0.833