Question

0.14 g of an element on combustion gives 0.28 g of its oxide. What is that element?

• A. Nitrogen
• B. Carbon
• C. Fluorine
• D. Sulphur

Answer 1

Answer: (D)

Eq. Weight of element $\text{= }\frac{\text{Weight}\text{of}\text{element}}{\text{Weiight}\text{of}\text{oxygen}}\times \text{ 8}$ $=\frac{0.14}{0.28-0.14}\times 8=8g^{g-equiv}$ Thus, the element should be suphur as its eq. Weight can be 8. $\underset{\begin{array}{c}\text{1}\text{mol}\\ \text{32}\text{g}\end{array}}{S}+O_2\to \underset{\begin{array}{c}\text{1}\text{mol}\\ \text{32 + (2}\times 16\text{)}\text{=}\text{64}\text{g}\end{array}}{S{O}_2}$ $\because$ 32 g sulphur, on combustion, gives oxide $=64g$ $\because$ 0.14 g sulphur, on combustion, will give oxide $=\frac{64\times 0.14}{32}=0.28g$ Hence, it is confirmed that the element is sulphur.