$\begin{aligned}
& \mathrm{NaI}{(\mathrm{aq})}+\mathrm{AgNO}{3(\mathrm{aq)}} \rightarrow \mathrm{AgI}_{(\mathrm{s})}+\mathrm{NaNO}_3(\mathrm{aq}) \
& \mathrm{M}, 20 \mathrm{ml} \text { excess } \
& 4.74 \mathrm{~g}
\end{aligned}$
Moles of I−in NaI= Moles of (I−)in AgI=2354.74
Moles of NaI=2354.74
Molarity [NaI]=235×0.024.74=1.008