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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Some Basic Concepts of Chemistry

JEE Main 2025 · 8 April · Evening Shift · Chemistry · Some Basic Concepts of Chemistry

20 mL of sodium iodide solution gave 4.74 g silver iodide when...

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

Correct answer

0

Step-by-step solution

NaI(aq)+AgNO3(aq)→AgI(s)+NaNO3(aq)&M,20ml excess &4.74g\begin{aligned}& \mathrm{NaI}_{(\mathrm{aq})}+\mathrm{AgNO}_{3(\mathrm{aq)}} \rightarrow \mathrm{AgI}_{(\mathrm{s})}+\mathrm{NaNO}_3(\mathrm{aq}) \& \mathrm{M}, 20 \mathrm{ml} \text{ excess } \& 4.74 \mathrm{ g}\end{aligned}\text{ Moles\text{ of } } I\mathrm{I}^{-} \text{in } \mathrm{NaI}=\text{ Moles of }\left(I\mathrm{I}^{-}\right) \text{in } \mathrm{AgI}=\frac{4.74}{235} \text{ Moles of } \mathrm{NaI}=\frac{4.74}{235} \text{ Molarity }[\mathrm{NaI}]=\frac{4.74}{235 \times 0.02}=1.008$
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