Question

A mixture of carbon monoxide and carbon dioxide is found to have a density of 1.7 g lit at s.t.p. the mole fraction of carbon monoxide is

Answer 1

Answer : The mole fraction of carbon monoxide is 0.368

Solution : Given,

Density of a mixture of gas = 1.7 g/liter

Molar mass of CO = 28 g/mole

Molar mass of $CO_2$ = 44 g/mole

At S.T.P (standard temperature & pressure),

Temperature = 273 K

Pressure = 1 atm

The formula used for molar mass of a mixture of gas in term of density is,

$M=\frac{\rho \times R\times T}{P}$              ...........(1)

where,

M = Molar mass of a mixture of gas

$\rho$ = Density of a mixture of gas

R = Gas constant = 0.0821 L atm/mole/K

T = temperature

P = Pressure

Now put all the given values in above formula(1), we get

$M=\frac{1.7g/L \times 0.0821 L atm/mole/K\times 273K}{1atm}$ = 38.102 g/mole

Let the mole fraction of CO is 'X' and the mole fraction of $CO_2$ will be (1-X).

The Average molar mass of a mixture is,

$M={\Sigma N\times M_i}$       ...........(2)

where,

N = mole fraction

$M_i$ = Molar mass of gas

Now put all the given values in above formula(2), we get

$38.102={28X+44(1-X)}$

By rearranging the terms, we get the value of 'X'

X = 0.368

Hence, the mole fraction of CO = X = 0.368