Question

Find the molar mass of phosphorus. Given the volume of phosphorus is $34.05\,ml$ and weight of phosphorus is $0.0625\,g$ at ${546^ \circ }C\& 1\,bar$ pressure.

Answer 1

Hint: We can calculate the molar mass of substance from the ideal gas equation.
The ideal gas equation is,
$PV = nRT$
Where,
 P is the pressure in the atmosphere.
V is the volume of gas in a liter.
n is the number of moles.
R is a universal gas constant.
T is the temperature.

Complete step by step answer:Given,
The volume of phosphorus is $34.05\,ml$.
The mass of phosphorus is $0.0625\,g$.
The given pressure$ = 1\,bar$
The given temperature$ = {560^ \circ }$
We know the value of R (a universal gas constant) is $0.08205\,L\,atm\,mo{l^{ - 1}}{K^{ - 1}}$.
The number of moles can be calculated from the ideal gas equation as,
$n = \dfrac{{PV}}{{RT}}$
We know that, the mole can be given by,
${\text{Mole = }}\dfrac{{mass\left( m \right)}}{{{\text{Molecular}}\,{\text{weight}}\left( M \right)}}$
Replace $n = \dfrac{m}{M}$
$\dfrac{m}{M} = \dfrac{{PV}}{{RT}}$
Substitute the known values in the above equation to find the molar mass.
$M = \dfrac{{\left( {0.08205\,L\,atm\,mo{l^{ - 1}}\,{K^{ - 1}}} \right)\left( {819\,K} \right)\left( {0.0625\,g} \right)}}{{\left( {0.98\,atm} \right)\left( {0.03405\,L} \right)}}$
$M = 125.863\,g/mol$
The molar mass of phosphorus is $125.863\,g/mol.$

Additional note:
Molar Mass:
The atomic mass of an element in grams contains Avogadro’s number of atoms and is defined as the molar mass of that element. To find the molar mass, we have to change the units of atomic mass from the atomic mass unit to grams.
For example, sulfur has an atomic mass of $32\,amu$ so one mole of sulfur has a molar mass of $32\,g$ and contains Avogadro’s number of atoms.


Note:
Always remember to convert pressure in the bar to the atmosphere unit.
$1\,bar = 0.98\,atm.$
Convert the volume of phosphorus in ml to L using the conversion,
$1\,L = 1000\,ml$
And also convert the temperature in Celsius to Kelvin using the conversion,
${0^ \circ }C = 273\,K$