Consider the following sequence of reactions:
i) \( \text{Mg}, \text{dry ether} \)
ii) \( \text{CO}_2, \text{H}_2\text{O}^+ \)
iii) \( \text{NH}_3, \Delta \)
11.25 mg of chlorobenzene will produce \( x \times 10^{-1} \) mg of product B. Given molar mass of C, H, O, N, and Cl as 12, 1, 16, 14, and 35.5 g mol\(^{-1}\), respectively, the value of \( x \) is:
ii) \( \text{CO}_2, \text{H}_2\text{O}^+ \)
iii) \( \text{NH}_3, \Delta \)
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Solution and Explanation
\[ \text{Moles of chlorobenzene} = \frac{11.25 \, \text{mg}}{1 \, \text{g/mol}} = \frac{11.25 \times 10^{-3}}{C_6H_5Cl} \]
Step 2: Calculate the corresponding amount of product B after completing the reaction. Final Conclusion: The value of \( x \) is 9.Top JEE Main Chemistry Questions
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