Question

The density of nitrogen under standard conditions is:
A. $2.5g/L$
B.\[1.25g/L\]
C. $0.625g/L$
D. $3.75g/L$

Answer 1

Hint: In order to solve this question, we need to know the volume occupied by nitrogen gas at standard conditions. Standard temperature and pressure (STP) are 273K and 1 atm respectively. At STP, the volume occupied by one mole of an ideal gas is $22.4L$.

Formula used:
$\rho = \dfrac{m}{V}$
Where, $\rho $ is density of one mole gas
$m$ is mass of one mole gas
$V$ is volume occupied by one mole of gas at standard temperature and pressure

Complete step by step answer:
We know that, at standard conditions one mole of dinitrogen gas will occupy $22.4L$ volume
$ \Rightarrow V = 22.4L$
One mole of nitrogen gas has mass = no. of moles of nitrogen x molecular mass of nitrogen
$ \Rightarrow m = 28g$
We know that density is given by,
$\rho = \dfrac{m}{V}$
Where, $\rho $is density of gas
$m$ is mass of gas
$V$ is volume occupied by gas
$
   \Rightarrow \rho = \dfrac{{28}}{{22.4}} \\
   \Rightarrow \rho = 1.25g/L \\
 $
Therefore, density occupied by nitrogen gas at standard conditions is $1.25g/L$

The correct option is B.

Note:
You can find out the volume of gas at standard conditions using the Ideal Gas Equation (We generally assume all gases to be ideal gases).
You may know, that the ideal gas equation is \[PV = nRT\]
Where, $P$ is pressure of gas,
$V$ is volume of gas,
$R$ is gas constant,
$T$is the absolute temperature
And $n$ is no. of moles of gas
STP (standard temperature and pressure) represents conventional conditions set by IUPAC with
$T = 273K$ and $P = 1atm$
Also, we know that $R = 0.0821L.atm/(Kmol)$
Now, putting these values in the ideal gas equation, volume for 1 mole of gas can be calculated as,
$
  PV = nRT \\
   \Rightarrow (1atm)(V) = (1mol)[0.0821L.atm/(Kmol)](273K) \\
   \Rightarrow V \approx 22.4L \\
$
Thus, we can conclude that volume occupied by one mole of a gas at standard conditions is found out to be $22.4Litres$.
The volume occupied by 1 mole of gas is known as the molar volume of the gas.
Thus, we can say that the molar volume of a gas at STP is $22.4Litres$.