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.
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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