$7 g$ of a sample of sodium chloride on treatment with excess of silver nitrate gave $14.35 g$ of $AgCl$. The percentage of $NaCl$ in the sample is

 The chemical reaction is as follows:

$NaCl+AgN{O}_3\to AgCl+NaN{O}_3$

From the reaction we can say that $1 \mathrm{mol} \mathrm{of} \mathrm{NaCl} \mathrm{gives} 1 \mathrm{mol} \mathrm{of} \mathrm{AgCl}.$

We can use the following relation: $\frac{W\left(NaCl\right)}{M\left(NaCl\right)}=\frac{W\left(AgCl\right)}{M\left(AgCl\right)}$

$\frac{W\left(NaCl\right)}{58}=\frac{14.35}{143.5}$

$W\left(NaCl\right)=5.845gm$

Percentage of NaCl= $\%\left(NaCl\right)=\frac{5.845}{7}\times 100=83.5\%$

Hence the correct option is (D).