JEE Main 2025 Chemistry question
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is M. (Nearest Integer Value)
Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol)
Answer: 1
Solution
Moles of AgI = 4.74/235 = 0.02. Moles of NaI = 0.02x×20/1000... solving: 0.02x=0.02, x=1M.