JEE Chemistry 2023 Question with Solution

Given: Mass of organic compound = $0.5 \, \text{g}$, carbon content = $60\%$.

Find: The value to fill in the blank for the mass of $CO_2$ produced on complete combustion.

Mass of carbon present in the compound:

$$0.5 \times 0.6 = 0.3 \, \text{g}$$

Moles of carbon:

$$\frac{0.3}{12} = \frac{0.5 \times 0.6}{12} = 0.025$$

On complete combustion, each mole of carbon gives one mole of $CO_2$. Therefore,

$$\text{moles of } CO_2 = 0.025$$

Mass of $CO_2$ produced:

$$0.025 \times 44 = \frac{0.5 \times 0.6}{12} \times 44 = 1.1 \, \text{g}$$

The question asks for _____ in _____ $\times 10^{-1} \, \text{g}$. Since

$$1.1 \, \text{g} = 11 \times 10^{-1} \, \text{g}$$

therefore, the required numerical value is $11$.

Given: The compound contains carbon, and on complete combustion carbon forms $CO_2$.

Find: The numerical value in the blank.

For combustion of an organic compound, carbon atoms convert to $CO_2$. Hence the number of moles of carbon present determines the number of moles of $CO_2$ formed.

Mass of carbon in $0.5 \, \text{g}$ sample:

$$0.5 \times \frac{60}{100} = 0.3 \, \text{g}$$

Moles of carbon:

$$\frac{0.3}{12} = 0.025 \text{ mol}$$

Thus,

$$\text{moles of } CO_2 = 0.025 \text{ mol}$$

Using molar mass of $CO_2 = 44 \, \text{g mol}^{-1}$,

$$\text{mass of } CO_2 = 0.025 \times 44 = 1.1 \, \text{g}$$

So the produced mass is $1.1 \, \text{g}$, which means the blank in _____ $\times 10^{-1} \, \text{g}$ is $11$.