Question

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 5.8-L bulb, then filled it with the gas at 1.00 atm and 21.0 ∘C and weighed it again. The difference in mass was 6.7 g. Identify the gas. Express your answer as a chemical formula.

Answer 1

Step-by-step explanation:

The given data is as follows.

Volume = 5.8 L, Pressure = 1.00 atm

T =
`21^(o)C` = (21 + 273) K = 294 K, mass = 6.7 g

R = 0.0821 Latm/mol K

According to the ideal gas equation, we will calculate the number of moles as follows.

PV = nRT

or, n =
`(PV)/(RT)`

Hence, putting the given values into the above formula as follows.

n =
`(PV)/(RT)`

=
`(1.00 atm * 5.8 L)/(0.0821 Latm/mol K * 294 K)`

= 0.24 mol

Now, calculate molar mass of the gas as follows.

Molar mass =
`\frac{mass}{\text{no. of moles}}`

=
`(6.7 g)/(0.24 mol)`

= 27.91 g/mol

or, = 28 g/mol (approx)

As the gas is diatomic so, mass of two atoms of gas X is 28 g/mol. And, mass of one atom of gas is as follows.

`(28 g/mol)/(2)`

= 14 g/mol

Hence, nitrogen atom has a mass of 14 g/mol. Therefore, we can conclude that the given diatomic gas is
`N_(2)`.