Q120
General ScienceChemistry
Formula of the Chloride of an element M is $\mathrm{MCl_4}$ and its Molecular weight is 154 (Cl = 35.5). The molecular weight of its oxide is
- a.64
- b.72
- c.44
- d.40
Answer: (C) 44
Molecular weight of $\mathrm{MCl_4} = M + 4 \times 35.5 = 154$, so the atomic weight of M is $154 - 142 = 12$, i.e. carbon, and the chloride is carbon tetrachloride. The formula $\mathrm{MCl_4}$ shows that M has valency 4, so its oxide is $\mathrm{MO_2}$, with molecular weight $12 + 2 \times 16 = 44$ (carbon dioxide). The other options are the weights of familiar oxides and act as traps: 64 is $\mathrm{SO_2}$, 72 is FeO and 40 is MgO. Two-step method for such questions: first find the atomic weight from the given compound, then use the valency to write the formula of the compound asked.