Question

What is the density of bromine vapour at STP?
(A) 2.5 g/L
(B) 2.9 g/L
(C) 3.6 g/L
(D) 4.9 g/L
(E) 7.1 g/L

Answer 1

Hint: Use the formula of ideal gas law but in a different form which is,
\[PM=DRT\]
The Molar mass of bromine gas is 160 g; at STP the temperature is zero degree Celsius and pressure is 1 atm.

Complete step by step solution:
We all know the ideal gas law which relates the pressure-volume temperature and the number of moles of gas to each other. In deriving this law, factors such as intermolecular forces of attraction and repulsion, the elastic and inelastic collision of gas particles with each other and also with the boundaries of the container they are kept in and the atomic volume of individual atoms are considered negligible. The law in its mathematical form is shown below-
\[PV=nRT\]
Where P is pressure, V is volume, n is the number of moles, R is the universal gas constant and T is the temperature of the gas taken into account. The constant R is actually a proportionality constant and its value remains the same for all conditions.
We can write the above equation in another form. The derivation is shown below:
\[\begin{align}
& PV=nRT \\
& \Rightarrow PV=\dfrac{m}{M}RT \\
\end{align}\]
Here, m is the mass of the gas present in the experiment and M is the molar mass of the same gas. Further,
\[\begin{align}
& PM=\dfrac{m}{V}RT \\
& \Rightarrow PM=DRT \\
\end{align}\]
Here, D is the density of the gas which is taken into account. So, we can also write the above equation as-
\[D=\dfrac{PM}{RT}\]
This is the equation that will give the answer to this question. We already have all the values with us. Putting them in their respective places, we get-
\[ \Rightarrow D=\dfrac{PM}{RT}=\dfrac{1\operatorname{atm}\times 160\operatorname{g}}{0.082\times 273.15\operatorname{K}}=7.14\operatorname{g}/L\]
The pressure is equal to 1 atm and the temperature is equal to 273.15 K at STP. The universal gas constant or R is equal to 0.082 L atm/ K.

So, the density of bromine gas under the given condition is 7.14 g/L. Therefore, the answer to this question will be option (E).

Note: The value of R is different if the units of the other quantities involved in the ideal gas equation are different. A student should memorise all the values and use them accordingly.