Question

Complete combustion of \( X \) g of an organic compound gave 0.25 g of CO\(_2\) and 0.12 g of H\(_2\)O. If the percent of carbon is 25% and of hydrogen is 4.8%, then \( X = \) _______ g (Nearest integer).}

• A. 273
• B. 274
• C. 273.5
• D. 227

Answer 1

The Correct Option is A

Step 1: Given information.
- Mass of CO\(_2\) = 0.25 g
- Mass of H\(_2\)O = 0.12 g
- Percent of carbon = 25%
- Percent of hydrogen = 4.8%
- Molar masses: \( C = 12 \), \( H = 1 \), \( O = 16 \)
Step 2: Finding the moles of carbon and hydrogen.
- Moles of carbon in CO\(_2\) = \( \frac{0.25}{44} \times 12 = 0.06818 \, \text{mol} \) (since CO\(_2\) has one mole of carbon per mole of CO\(_2\)) - Moles of hydrogen in H\(_2\)O = \( \frac{0.12}{18} \times 2 = 0.01333 \, \text{mol} \) (since H\(_2\)O has two moles of hydrogen per mole of H\(_2\)O)
Step 3: Using the given percentages to find \( X \).
- Carbon content \( = 25% \), so the total mass of the compound is \( X = \frac{0.06818 \times 12}{0.25} = 273 \, \text{g} \).
Step 4: Conclusion.
Therefore, the value of \( X \) is 273 g.
Final Answer: (A) 273