20 mL of sodium iodide solution |LIQUID SOLUTIONJEEMAIN2023

JEE MAIN 2025

08-04-2025 S2

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: $\mathrm{Na}=23, \mathrm{I}=127, \mathrm{Ag}=108, \mathrm{~N}=14, \mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ )

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Solution

$$ \mathrm{NaI}_{(\mathrm{aq})}+\mathrm{AgNO}_{3(\mathrm{aq})} \rightarrow \mathrm{AgI}_{(\mathrm{s})}+\mathrm{NaNO}_{3(\mathrm{aq})} $$ M, 20 ml excess $\quad 4.74 \mathrm{~g}$ Moles of $\mathrm{I}^{-}$in $\mathrm{NaI}=$ Moles of $\left(\mathrm{I}^{-}\right)$in $\mathrm{AgI}=\frac{4.74}{235}$ Moles of NaI $=\frac{4.74}{235}$ Molarity $[\mathrm{NaI}]=\frac{4.74}{235 \times 0.02}=1.008$

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JEE Main

Chemistry

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