JEE Chemistry 2025 Question with Solution

Given: Mass of organic compound = $0.5 \, \text{g}$, mass of $CO_2$ obtained = $1.46 \, \text{g}$.

Find: Percentage of carbon in the compound.

During combustion, all the carbon present in the compound is converted into $CO_2$.

Using the molar mass ratio:

$$\text{Mass of C in compound} = \frac{1.46}{44} \times 12$$

Now calculate the percentage of carbon:

$$\%\text{ of C} = \frac{1.46}{44} \times \frac{12}{0.5} \times 100$$ $$= 79.63$$

Rounding to the nearest integer, the percentage of carbon is $80\%$. Therefore, the required answer is 80.

Given: Mass of compound = $0.5 \, \text{g}$, mass of $CO_2$ = $1.46 \, \text{g}$.

Find: Percentage of carbon in the compound.

Step 1: Calculate moles of $CO_2$ produced.

$$\text{Moles of } CO_2 = \frac{1.46 \, \text{g}}{44 \, \text{g/mol}} \approx 0.0332 \, \text{mol}$$

Step 2: Each mole of $CO_2$ contains one mole of carbon, so moles of carbon = $0.0332 \, \text{mol}$.

Mass of carbon:

$$\text{Mass of carbon} = 0.0332 \times 12 \approx 0.3984 \, \text{g}$$

Step 3: Percentage of carbon in the compound:

$$\text{Percentage of carbon} = \left(\frac{0.3984}{0.5}\right) \times 100 \approx 79.68\%$$

Rounding to the nearest integer gives $80\%$.

Therefore, the required answer is 80.