Question

Given,8.2 L of an ideal gas weighs 9.0 g at 300 K and 1 atm pressure. The molecular mass of gas is:-
(a) 9
(b) 27
(c) 54
(d) 81

Answer 1

Hint: As we know that the ideal gas law is a good approximation of the behavior of various gases under certain conditions. It is a combination of the empirical Boyle's law, Charles's law, Gay-Lussac's law and Avogadro’s law. So here we have to calculate the molecular mass of the gas using the ideal gas equation.
Formula used:
We will use ideal gas equation (equation of state):
$PV=nRT$
$n=\dfrac{W}{M}$
where,
P = pressure of the gas
V =volume of gas it occupies
n = number of moles of gas present in the solution or sample
R = universal gas constant
T = absolute temperature of the gas
W= given mass
M =molecular mass of the gas

Complete answer:
Let us first discuss the ideal gas equation as follows:-
Ideal gas equation: It is the empirical relationship between volume, temperature, pressure and the amount of gas combined together into the ideal gas law which can be mathematically written as follows:-
$PV=nRT$
where,
P = pressure of the gas
V =volume of gas it occupies
n = number of moles of gas present in the solution or sample
R = universal gas constant, equal to $0.0821\dfrac{atm\cdot L}{mol\cdot K}$
T = absolute temperature of the gas
-Calculation of the moles of the gas as follows:-
The values given in the question are:-
P = 1 atm
V = 8.2L
T = 300K
On substituting all the values in ideal gas equation, we get:-
\[\begin{align}
  & \Rightarrow PV=nRT \\
 & \text{Rearrange the formula:-} \\
 & \Rightarrow \text{n =}\dfrac{PV}{RT} \\
 & \Rightarrow \text{n =}\dfrac{1atm\times 8.2L}{300K\times 0.0821\dfrac{atm\cdot L}{mol\cdot K}} \\
 & \Rightarrow \text{n =}0.3329moles \\
\end{align}\]
-Calculation of molecular mass of gas using mole concept:-
The given mass (W) of gas is = 9.0g
On substituting this value in$n=\dfrac{W}{M}$, we get:-
$\begin{align}
  & \Rightarrow 0.3329moles=\dfrac{9.0g}{M} \\
 & \Rightarrow M=\dfrac{9.0g}{0.3329moles} \\
 & \Rightarrow M=27g/mol \\
\end{align}$

Therefore, the molecular mass of the gas is: (b) 27g/mol

Note:
-Remember to change all the values with respect to universal gas constant and then use them in the formula. This will help you to get an accurate answer with minimum error.
Also there are different values of universal gas constant as shown below:-
-8.314$J\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}$
-8.314${{m}^{3}}\cdot Pa\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}$
-2 $cal\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}$ -8.205${{m}^{3}}\cdot atm\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}$
-0.082$L\cdot atm\cdot {{K}^{-1}}\cdot mo{{l}^{-1}}$