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* 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= (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 = (given weight)/(Molar mass)`

`n = (35* 10^(-3) g)/(28 g/mol) = 1.25* 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= (1.25* 10^(-3) mole* 8.314 Pa m^(3)/K mol* 298.15 K)/(101325Pa)`

`V = 3.06* 10^(-5) m^{^(3)}`

Now converting
`m^{^(3)}` to ppmv as:

`V = 3.06* 10^(-5) \frac{m^{^(3)} CO}{m^{^(3)} air}* 10^(6)`

`V = 30.6 ppm_v`

b. Percent by volume is calculated as:

`Volume percent = (Volume of CO)/(Volume of CO+Volume of air)* 100`

`Volume percent = (3.06* 10^(-5) m^(3))/(3.06* 10^(-5) m^(3)+ 1 m^(3))* 100`

`Volume percent = 3.06* 10^(-3)`%.