Question

$34.05ml$ of phosphorus vapour weighs $0.0625g$ at ${{546}^{o}}C$ and $0.1bar$ pressure. What is the molar mass of phosphorus?

Answer 1

To solve this question, consider the formula of gas equation which considers the four variables i.e. volume, pressure, temperature and number of moles. So, in the question volume, pressure, temperature and weight of the gas is already given and you know the number of moles equals the given weight divided by its molar mass. Now, hope you can get your answer.

Complete step by step answer:
Given that,
Volume of phosphorus vapour is $34.05ml$which can be written as $34.05\times {{10}^{-3}}L=34.05\times {{10}^{-3}}d{{m}^{3}}$.
Pressure of the gas is $0.1bar$.
Temperature of the gas is ${{546}^{o}}C$ which can also be written as $(546+273)K = 819K$.
Mass of the vapour gas is $0.0625g$.
So, to get molar mass of the gas we have to use the formula of gas equation i.e.
$PV=nRT$
Where, P is the pressure of the gas,
V is the volume of the gas,
T is the temperature,
n is the number of moles, and
R is the universal gas constant which equals to $0.08314L-Bar/K-mol$

So, this equation can also be written as
$PV=\dfrac{MassgivenRT}{Molar Mass}$
As the number of moles equals the mass given of a substance divided by its molar mass.
So, now molar mass will be equal to,
$Molar mass=\dfrac{MassgivenRT}{PV}$
Now, by substituting the given values in the above equation, we will get the molar mass of the gas.
So, $Molar mass=\dfrac{0.0625\times 0.08314\times 819}{1\times 34.05\times {{10}^{-3}}}=124.75g/mol$
Hence, the molar mass of phosphorus vapour will be $124.75g/mol$.

Note: The ideal gas equation is the combination of empirical laws like Charles’ law, Boyle’s law, Gay-Lussac’s law and Avogadro’s law. While solving such problems, take into consideration the units of the variable given and whenever required convert the units of the variables to simplify the calculation.