Question

34.05 mL of phosphorus vapour weighs 0.0625 g at 546${}^\circ C$ and 1.0 bar pressure. What is the molar mass of phosphorus?

Answer 1

The derivation of the Ideal Gas Equation is helpful in determining the molar mass of an unknown gas .We have the values of pressure, temperature, volume of phosphorus vapour and the weight is also given. By substituting these values in the equation we could find the molar mass of phosphorus.

Complete step by step solution:
- As we know, molar mass is the mass of one mole of a compound or an element. They are expressed in units of grams per mole and are also referred to as molecular weight.
- The Ideal Gas Law is an equation of state for a gas, which describes the relationship among the four variables such as moles of gas (n), temperature (T), pressure (P), and volume (V). The modified form of this Ideal Gas equation involves the molecular weight (MW) and the mass (m) instead of volume (V) and moles (n). The gas equation can be mathematically represented as follows
$PV=nRT$
Where P is the pressure (1.0 bar)
V is the volume (34.05 mL or$34.05\times {{10}^{-3}}d{{m}^{3}}$)
n is the number of moles
T is the temperature (546${}^\circ C$ or 819 K)
R is the gas constant (0.083 bar $d{{m}^{3}}{{K}^{-1}}mo{{l}^{-1}}$)
The above gas equation can be modified in terms of molar mass as follows
$M=\dfrac{wRT}{PV}$
Where M is the molar mass which we need to find and w is the mass of the gas (0.0625 g). Let’s substitute the given values in the above equation
$M=\dfrac{0.0625\text{ }\times 0.08314\times 819}{1\times 34.05\times {{10}^{-3}}}=124.98g$

Therefore the molar mass of phosphorus is 124.98 g

Note: Keep in mind that the ideal gas equation has many limitations. Ideal gas equation holds well as long as the density is kept low and is applicable for single gas or even a mixture of multiple gasses in which ‘n’ will stand for the total moles of gas particles in the given mixture.