A \(512 cm ^{3}\) sample of a gas weighed \(1.236 g\) at \(20^{\circ} C\) and a pressure of one atmosphere. The relative molecular mass of the gas is \(\left[R=8.314 K ^{-1} mol ^{-1} ; 1 atm =101,325 Jm ^{-3}\right]\).
Explanation
Assuming the ideal gas behaviour i.e.
\begin{aligned}
&P V=n R T, \\
&P V=\frac{m R T}{M} ;\left(n=\frac{m}{M}\right)
\end{aligned}
Where \(P\) - pressure,
\(V\) - volume, \(m\) - mass,
\(M\) - molecular weight
\(M=\frac{m R T}{P V}\)
\(R=8.314 Jk ^{-1} mol ^{-1}\),
\(T=20+273.15=293.15 K\)
\(P=1 atm =101325 Jm ^{-3}\)
\(V=512 cm ^{3}=512 \times 10^{-6} m ^{3}\),
\(M=\frac{1.236 \times 8.314 \times 293.15}{101225 \times 512 \times 10^{-6}}=58.07\)
