Question

At its boiling point $\left(\mathbf{280}\mathbf{°}\mathbf{C}\right)$and at atmospheric pressure, phosphorus has a gas density of$\mathbf{2}\mathbf{.}\mathbf{7}{\mathbf{gL}}^{\mathbf{-}\mathbf{1}}$ . Under the same conditions, nitrogen has a gas density of$\mathbf{0}\mathbf{.}\mathbf{62}{\mathbf{gL}}^{\mathbf{-}\mathbf{1}}$ . How many atoms of phosphorus are there in one phosphorus molecule under these conditions?

Short answer

The number of atoms in one phosphorus molecule is 4.

Step-by-step solution

Number of atoms

The formula used to calculate the number of atoms is:

$\mathbf{Z}\mathbf{=}\frac{\mathbf{Molarmassofelement}}{\mathbf{Originalmassofelement}}$

The molar mass can be obtained as:

$\mathbf{W}\mathbf{=}\frac{\mathbf{DRT}}{\mathbf{P}}$

The number of atoms of phosphorus present in one phosphorus molecule under these conditions

Let the molar mass of phosphorus at the given conditions be W.

For ideal gas it is given that:

$\text{W =}\frac{\text{DRT}}{\text{P}}$

Where,

D represent the density of phosphorus gas which is equal to 2.7 g/L.

R represent the universal gas constant which is equal to 0.0821 atm.L/molK

T represent the temperature which is equal to $\text{280°C}$.

P represent the atmospheric pressure which is equal to 1 atm.

The data can be substituted in the equation as:

$$\begin{gathered} \text{W =}\frac{\text{DRT}}{\text{P}} \\ \text{=}\frac{{\text{2.7gL}}^{\text{- 1}}\text{×0.0821atmL/mol.K×553K}}{\text{1atm}} \\ \text{=}\frac{\text{122.58g}}{\text{mol}} \end{gathered}$$

The original mass of phosphorus is 31 g/mol.

Let the number of atoms in one phosphorus molecule be Z.

The value of Z can be found as:

$$\begin{gathered} \text{Z =}\frac{\text{Molar mass of phosphorus}}{\text{Original mass of phosphorus}} \\ \text{=}\frac{{\text{122.5835 g mol}}^{\text{- 1}}}{{\text{31g mol}}^{\text{- 1}}} \\ \text{= 4} \end{gathered}$$

The number of atoms in one phosphorus molecule is 4.