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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