## Answers

Answer: Given that, Molarity of KI = 0.209 M Mass of Cu CNO3)2 = 2.43 g Molari mass of Cu CNO3) = 187.56 g/mol So, No. of moles of Cu(NO3)2 - given mass - 2.43 g - Molat mass moles of Cu (NO3)2 = 0.013 males The balance reaction is 2 Cu(NO3), caq) + 4 KT cag) → 2 Cut cap) +T26 + 4 kNO, (01) 187.56 g/mol 2 male of eu (NO3), react with 4 mL of KI So, 0.013 moles of cu(NO3)2 will react with D = 24 x 0.013 molas = 0.026 moles of kI Now, Volume of KI Volume of KI = us Monty 126 no. of males of KI Molarity of KS 0.026 o dog = 0.1244 L V

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