Question

At standard pressure and temperature conditions, calculate the density of a gas whose molecular weight is\[45g/mol\].
A. 2
B. 22.4
C. 11.2
D. 1000

Answer 1

Hint: Density is defined as mass per unit volume of a substance under specific conditions of temperature and pressure. The density of the gas is equal to its mass divided by the volume. The Density varies with temperature and pressure.
\[\;The{\text{ }}density{\text{ }}of{\text{ }}gas = molecular{\text{ }}wt.{\text{ }}of{\text{ }}metal{\text{ }}/volume\]

Complete Step by step answer: The original ideal gas law uses the formula PV = nRT, the density version of the ideal gas law is
$PM = \rho RT$
The density of gas formula is mathematically expressed as-
\[\]$\rho = P(MW)/RT$
Where in,
$\rho $ – ideal gas density in Kg/l
P – refers to the pressure of the gas in atm
R – universal gas constant \[\left( {8.32{\text{ }}Joule} \right)\]
T – absolute temperature of the gas in Kelvin
MW – Molecular weight
Given:
At standard pressure and temperature conditions (STP)
Molecular weight (MW) = \[45{\text{ }}g/mol\]
T = \[273{\text{ }}K\]
P = \[1{\text{ }}atm\]
R= \[0.0821{\text{ }}atm.L{\text{ }}/Mol.K\]
Now we need to insert the values we know.
$\rho = P(MW)/RT$
$\Rightarrow \rho $=\[45{\text{ }}X{\text{ }}1/0.0821{\text{ }}X{\text{ }}273\] , \[ = {\text{ }}45/22.41\] $\rho $= \[2.008{\text{ }}g/l{\text{ }},\;\;\]
$\Rightarrow \rho $ = \[2.0{\text{ }}g/l\]

The correct option is (A).

Note: we can change the density of an element or compound by changing either the temperature or the pressure. Furthermore, increasing the pressure always increases the density of a substance. On the other hand, increasing the temperature generally decreases the density but there are certain exceptions to it. For example, in the case of water, its density increases between its melting point\[{0^o}C\] and \[{4^o}C\].