\(0.79 g\) of a gas at s.t.p. occupied a volume of \(250 cm ^{3}\). What is the relative molecular mass of the gas? (G.M.V. at s.t.p. \(=22.4 dm ^{-3}\) )
A. 17 B. 32 C. 64 D. 71
Correct Answer: D
Explanation
No of mole \(=\frac{\text { Mass of gas }}{\text { Molar mass of gas }}\) i.c. \(\eta=\frac{m}{M} ;\) Also, no of mole of gas \(=\frac{\text { Vol. of gas at s.t.p. }}{\text { Molar vol. of gas }}\) i.e. \(\eta=\frac{V_{\text {s.p. }}}{22.4 dm ^{3} / mol }\) connecting the two expressions, we have \begin{aligned} &\frac{m}{M}=\frac{V_{\text {s.t.p. }}}{22.4 dm ^{3} / mol } \\ &M=\frac{m \times 22.4 dm ^{3} / mol }{V_{\text {st.p. }}} \\ &=\frac{0.79 g \times 22.4 dm ^{3} / mol }{250 \times 10^{-3} dm ^{3}} \\ &=70.784 g / mol \approx 71 gmol \\ &\text { R.M.M }=71 g \end{aligned}