Answer:
The identity of the gas is Cl₂ (chlorine)
Explanation:
STP conditions are:
1 atm → Pressure
273.15 K → T°
So, we must use the Ideal Gas Law to get the moles.
Before that, we will think density as data where 3.164 g of diatomic gas are contained in 1 L of volume.
P.V = n . R . T
1 atm . 1 L = n. 0.082 L.atm/mol.K . 273.15K
1 L.atm / (0.082 L.atm/mol.K . 273.15K) = n
0.0446 mol = n
This quantity of diatomic gas, are 3.164 g so the molar mass will be:
Mass / mol = molar mass
3.164 g / 0.0446 mol = 70.9 g/m
The element (a diatomic molecule), which has that molar mass in the periodic table is the Cl₂.
1 Cl = 35.45 g/m
Cl₂. = 70.9 g/m
The identity of the diatomic gas is Chlorine (Cl)
From the question, we are to determine the identity of a gas that has a density of 3.164 g/L at stp.
To determine the identity of the diatomic gas, we will determine the mass of 1 mole of the gas
From the formula
[tex]Density = \frac{Mass}{Volume }[/tex]
Then,
Mass = Density × Volume
Also, at standard temperature and pressure (STP), 1 mole of any gas will occupy a volume of 22.4 L
∴ Mass of the diatomic gas = 3.164 × 22.4
Mass of 1 mole of the diatomic gas = 70.8736 g
Now, we will determine the atomic mass of the gas
Since, the gas is a diatomic gas,
The atomic mass of the gas = [tex]\frac{70.8736}{2}[/tex]
The atomic mass of the gas = 35.4368 g/mol
The atomic mass of the gas ≅ 35.44 g/mol
This is reasonably close to the atomic mass of chlorine ( At. mass of Cl = 35.453 g/mol)
Hence, the identity of the diatomic gas is Chlorine (Cl)
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