Question

What is the molar mass of a gas that has a density of $1.02g{L^{ - 1}}$ at $0.990$ atm pressure and ${37^ \circ }C$$.$

Answer 1

Hint :To solve this question $,$ we must know the ideal gas law $:$ $PV = nRT$$.$ Since the number of moles $n$ is not provided directly in the question $,$ we must solve it using density $\rho $$.$Density is mass of unit volume of a substance that is mass divided by volume and number of moles is given mass divided by molar mass of the gas $.$ The value of universal gas constant will be $0.0821atmLmo{l^{ - 1}}{K^{ - 1}}$$.$

Complete Step By Step Answer:
We know the ideal gas equation $,$$PV = nRT$$,$ where P is the pressure of the gas $,$ V is the volume of the gas $,$ n is the number of moles of the gas $,$ R is the universal gas constant and T is the temperature of the gas in Kelvin $.$
Given $,$ P $ = $$0.990$atm
T $ = $ ${37^ \circ }$C $ = $ $37 + 273.15 = 310.15K$
R $ = $ $0.0821atmLmo{l^{ - 1}}{K^{ - 1}}$
Since, volume and number of moles of the gas is not given $,$ we have to find it out with the help of density equation $.$
Number of moles $,$ $n = \dfrac{{givenmass(m)}}{{Molarmass(M)}}$
Density $,$ $\rho = \dfrac{{mass(m)}}{{Volume(V)}}$
Rewriting the ideal gas equation, $PV = \dfrac{m}{M}RT$
$M = \dfrac{m}{V}\dfrac{{RT}}{P}$
$M = \rho \dfrac{{RT}}{P}$
Now, substituting the given values, we get the molar mass of the gas as $:$
$M = \dfrac{{1.02g{L^{ - 1}} \times 0.0821atmLmo{l^{ - 1}}{K^{ - 1}} \times 310.15K}}{{0.990atm}}$ $ = $ $26.23gmo{l^{ - 1}}$
Therefore $,$ the molar mass of the gas is $26.23gmo{l^{ - 1}}$ $.$

Note :
 The values of universal gas constant can vary according to the units of pressure $,$ volume given in the question $.$ Here $,$ since the pressure was in atm and volume in L $,$we substituted R value as $0.0821atmLmo{l^{ - 1}}{K^{ - 1}}$$.$ For e g $,$ if the pressure was given in Joules and volume in L $,$ R value will be $8.314Jmo{l^{ - 1}}{K^{ - 1}}$ $.$ Also, while substituting temperature in the ideal gas law equation $,$ it should be converted to Kelvin $(K)$ $.$