\(0.06 g\) of a hydrocarbon occupies \(32 cm ^{3}\) at S.T.P. Its formula is ____________
A. \(C _{3} H _{6}\) B. \(C _{2} H _{2}\) C. \(C _{2} H _{4}\) D. \(C _{2} H _{6}\)
Correct Answer: A
Explanation
\begin{aligned} \text { No of mole } &=\frac{\text { Volume at s.t.p. }}{22400 cm ^{3}} \\ &=\frac{\text { mass of hydrocarbon }}{\text { Molar mass of hydrocarbon }} \\ &=\frac{\text { Volume at s.t.p. }}{22400 cm ^{3} / mol } \\ \frac{0.06 g }{M} &=\frac{32 cm ^{3}}{22400 cm ^{3} / mol } \\ M &=\frac{0.06 g \times 22400 cm ^{3} mol ^{-1}}{32} \\ & \equiv[(12 \times 3)+(1 \times 6)] g / mol ^{3} \\ &=42 g / mol \\ & \therefore T hydrocarb \end{aligned} \(\therefore\) The hydrocarbon has the formular \(C _{3} H _{6}\)
