Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2025 — 8 April, Evening Shift — Question 19

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

Answer: 1

Numerical answer — enter this value.

Let molarity of Nal solution be $\times \mathrm{M}$

$\mathrm{NaI}+\mathrm{AgNO}_{3} \rightarrow \mathrm{AgI}+\mathrm{NaNO}_{3}$

Moles of Agl formed $=\frac{4.74}{235}=0.02$

Moles of $\mathrm{NaI}=\frac{20 \times x}{1000}=0.02 x$

$0.02 x=0.02$

$x=1$

$\therefore \quad$ Molarity of Nal solution $=1 \mathrm{M}$

Answer key and solution verified before publishing.

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