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}…

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

Aicharya · Accepted Answer

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

20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate…

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