250 mg of an organic compound gives 141 mg of AgBr by Carius method. The percentage of bromine in the compound is (Atomic weight of Ag = 108; Br = 80)

Chemistry
250 mg of an organic compound gives 141 mg of AgBr by Carius method. The percentage of bromine in the compound is (Atomic weight of Ag = 108; Br = 80)

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Answer

236.18\ mg

Step 1: प्रति अणु द्रव्यमान (Molar mass) ज्ञात करें।

KAg(CN)2 ⁣:K=39, Ag=108, C2=2×12=24, N2=2×14=28KAg(CN)_2 \colon \quad K = 39, \ Ag = 108, \ C_2 = 2 \times 12 = 24, \ N_2 = 2 \times 14 = 28

39+108+24+28=199 g/mol39 + 108 + 24 + 28 = 199\ g/mol

AgBr ⁣:108+80=188 g/molAgBr \colon \quad 108 + 80 = 188\ g/mol

Step 2: प्रतिक्रिया लिखें।

\ceKAg(CN)2+KBr−>AgBr+2KCN\ce{KAg(CN)2 + KBr -> AgBr + 2KCN}

1 mol \ceKAg(CN)2\ce{KAg(CN)2} से 1 mol \ceAgBr\ce{AgBr} प्राप्त होता है।

Step 3: \ceKAg(CN)2\ce{KAg(CN)2} के मोल ज्ञात करें।

m=250 mg=0.250 gm = 250\ mg = 0.250\ g

n=0.250 g199 g/mol=0.250199 moln = \frac{0.250\ g}{199\ g/mol} = \frac{0.250}{199}\ mol

Step 4: \ceAgBr\ce{AgBr} का द्रव्यमान ज्ञात करें।

m\ceAgBr=0.250199×188 gm_\ce{AgBr} = \frac{0.250}{199} \times 188\ g

=188199×250 mg= \frac{188}{199} \times 250\ mg

250×188=47000250 \times 188 = 47000

47000÷199=236.18 mg47000 \div 199 = 236.18\ mg

उत्तर: 236.18 mg\boxed{236.18\ mg}

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प्रति अणु द्रव्यमान (Molar mass) ज्ञात करें। KAg(CN)_2 K = 39, \ Ag = 108, \ C_2 = 2 × 12 = 24, \ N_2 = 2 × 14 = 28 39 + 108 + 24 + 28 = 199\ g/mol AgBr 108 + 80 = 188\ g/mol Step 2: प्रतिक्रिया लिखें। KAg(CN)2 + KBr -> AgBr + 2KCN 1 mol KAg(CN)2 से 1…

250 mg of an organic compound gives 141 mg of AgBr by Carius method. The percentage of bromine in the compound is (Atomic weight of Ag = 108; Br = 80)
Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

ScanSolve AI Answer
Step 1: प्रति अणु द्रव्यमान (Molar mass) ज्ञात करें। KAg(CN)_2 K = 39, \ Ag = 108, \ C_2 = 2 × 12 = 24, \ N_2 = 2 × 14 = 28 39 + 108 + 24 + 28 = 199\ g/mol AgBr 108 + 80 = 188\ g/mol Step 2: प्रतिक्रिया लिखें। KAg(CN)2 + KBr -> AgBr + 2KCN 1 mol KAg(CN)2 से 1 mol AgBr प्राप्त होता है। Step 3: KAg(CN)2 के मोल ज्ञात करें। m = 250\ mg = 0.250\ g n = 0.250\ g199\ g/mol = (0.250)/(199)\ mol Step 4: AgBr का द्रव्यमान ज्ञात करें। m_AgBr = (0.250)/(199) × 188\ g = (188)/(199) × 250\ mg 250 × 188 = 47000 47000 ÷ 199 = 236.18\ mg उत्तर: 236.18\ mg