Question

What is the concentration in (a) ppmv, and (b) percent by volume, of carbon monoxide (co) that has a concentration of 35 mg/m3? the problem conditions are temperature = 25oc and pressure = 1.0 atm?

Answer 1

Molar mass of carbon monoxide ($CO$) = $12+16 = 28 g/mol$

Concentration of carbon monoxide ($CO$) = $35 mg/m^{3}$ (given)

Since, $1 mg = 10^{-3} g$

So, the concentration of carbon monoxide ($CO$) = $35\times 10^{-3} g/m^{3}$

a. ppmv stands for parts per million by volume is defined as the volume of a substance dissolved in one million parts per volume of the liquid.

The formula used for determining the volume is:

$PV = nRT$

$V= \frac{nRT}{P}$  -(1)

where,

$V$ is volume, $n$ is number of moles, $R$ is universal gas constant, $T$ is temperature and $P$ is pressure.

For determining number of moles, $n$:

$n = \frac{given weight}{Molar mass}$

$n = \frac{35\times 10^{-3} g}{28 g/mol} = 1.25\times 10^{-3} mole$

Temperature, $T = 25^{o}C = 25+273.15K = 298.15 K$

Pressure, $P$ = $1 atm =101325 Pa$

Substituting the values in formula (1):

$V= \frac{1.25\times 10^{-3} mole\times 8.314 Pa m^{3}/K mol\times 298.15 K}{101325Pa}$

$V = 3.06\times 10^{-5} m^{^{3}}$

Now converting $m^{^{3}}$ to ppmv as:

$V = 3.06\times 10^{-5} \frac{m^{^{3}} CO}{m^{^{3}} air}\times 10^{6}$

$V = 30.6 ppm_v$

b. Percent by volume is calculated as:

$Volume percent = \frac{Volume of CO}{Volume of CO+Volume of air}\times 100$

$Volume percent = \frac{3.06\times 10^{-5} m^{3}}{3.06\times 10^{-5} m^{3}+ 1 m^{3}}\times 100$

$Volume percent = 3.06\times 10^{-3}$%.