In the Carius method, of an organic compound gave of . The percentage of bromine in the compound is: — Practical Organic Chemistry and Purification Chemistry Question
Question
In the Carius method, $0.15\text{ g}$ of an organic compound gave $0.12\text{ g}$ of $AgBr$. The percentage of bromine in the compound is: (Atomic mass $Ag=108, Br=80$)
Answer: C
💡 Solution & Explanation
$\%Br = (80 / 188) \times (Mass\ of\ AgBr / Mass\ of\ cmpd) \times 100$. Thus, $(80 / 188) \times (0.12 / 0.15) \times 100 = 34.04\%$.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes