The main component of table salt is $\mathrm{NaCl}$, and the chlorine content in table salt can be determined using the $\bf{Mohr}$ method. The operational steps are as follows:

($1$) Accurately weigh $1.3567 \mathrm{ g}$ of table salt, dissolve it in distilled water, and dilute to $250 \mathrm{~mL}$ in a volumetric flask.

($2$) Transfer $25.00 \mathrm{~mL}$ of this solution to an Erlenmeyer flask, add $6$ drops of $50.0 \mathrm{~g} / \mathrm{L}$ $\mathrm{K_2 Cr_2 O_7}$ solution, and titrate with $0.1000 \mathrm{~mol} \cdot \mathrm{L}^{-1}$ $\mathrm{AgNO_3}$ solution to the endpoint, consuming $22.61 \mathrm{~mL}$. The blank experiment consumed $0.05 \mathrm{~mL}$ of $\mathrm{AgNO_3}$ solution.

Assume that the impurity ions in the table salt do not affect the reaction, and assume that $\mathrm{Cr_2O_7^{2-}}$ is completely converted to $\mathrm{CrO_4^{2-}}$.

$$ K_{\mathrm{sp}}(\mathrm{AgCl})=1.8 \times 10^{-10} ; K_{\mathrm{sp}}\left(\mathrm{Ag_2CrO_4}\right)=1.2 \times 10^{-12} $$

Then the following statement that is correct is:
食盐的主要成分为 $\mathrm{NaCl}$, 可利用 $\bf{Mohr}$ 法测定食盐中的氯含量。其操作步骤如下:

($1$)准确称取 $1.3567 \mathrm{ g}$ 食盐,用蒸馏水溶解,定容至 $250 \mathrm{~mL}$ 容量瓶。

($2$)移取 $25.00 \mathrm{~mL}$ 该溶液锥形瓶中, 加入 $6$ 滴 $50.0 \mathrm{~g} / \mathrm{L}  $ $\mathrm{K_2 Cr_2 O_7}$ 溶液, 以 $0.1000 \mathrm{~mol} \cdot \mathrm{L}^{-1}$
的 $\mathrm{AgNO_3}$ 溶液滴定至终点, 消耗了 $22.61 \mathrm{~mL}$ 。空白试验消耗 $\mathrm{AgNO_3}$ 溶液 $0.05 \mathrm{~mL}$ 。

设食盐中的杂质离子不对反应产生影响, 并且假设 $\mathrm{Cr_2O_7^{2-}}$ 完全转化为 $\mathrm{CrO_4^{2-}}$。 

$$ K_{\mathrm{sp}}(\mathrm{AgCl})=1.8 \times 10^{-10} ; K_{\mathrm{sp}}\left(\mathrm{Ag_2CrO_4}\right)=1.2 \times 10^{-12} $$  

则下列说法正确的是：
All other options are incorrect
The sodium chloride content in table salt is $95.66\%$.
Assume the titration endpoint is exactly the stoichiometric point. The volume of each drop of $50.0 \mathrm{~g} / \mathrm{L}$ $\mathrm{K_2 Cr_2 O_7}$ solution is $0.05 \mathrm{~mL}$, at which point $\mathrm{Ag_2CrO_4}$ precipitate forms.
Under the condition of option $\bf{C}$, the silver ion concentration in the solution is $8.52 \times 10^{-6} \mathrm{~mol} \cdot \mathrm{L}^{-1}$ at this time.
To ensure that the endpoint error of this titration is $0$, the volume of indicator solution to be added should be $0.94 \mathrm{mL}$.
The method has smaller errors and better results when titrating $\mathrm{I}^{-}$ and $\mathrm{SCN}^{-}$.
其他选项均不正确
食盐中氯化钠的含量为$95.66\%$。
设滴定终点恰为化学计量点, 每滴 $50.0 \mathrm{~g} / \mathrm{L}$ $\mathrm{K_2 Cr_2 O_7}$ 溶液的体积为 $0.05 \mathrm{~mL}$，此时有 $\mathrm{Ag_2CrO_4}$ 沉淀生成。
在选项$\bf{C}$的条件下，此时溶液中银离子浓度为 $8.52 \times 10^{-6} \mathrm{~mol} \cdot \mathrm{L}^{-1}$。
为使得该滴定的终点误差为 $0$,应加入指示剂溶液的体积$0.94 \mathrm{mL}$。
该方法滴定 $\mathrm{I}^{-}$与 $\mathrm{SCN}^{-}$时误差较小，效果较好。
Sodium chloride content calculation:

First, find the amount of $\mathrm{AgNO_3}$ consumed by titration, and then obtain the amount of $\mathrm{Cl^-}$. Combined with solution dilution, extrapolate back to the amount of $\mathrm{Cl}$ in the original salt sample, and then calculate the sodium chloride content. The calculation is as follows:

$$\omega=\frac{0.1000 \times(22.61-0.05) \times 10^{-3} \times(35.45+22.99) \times 10} {1.3567}=97.18 \%$$

<Checkpoint>[Therefore, the sodium chloride content is $97.18 \%$. $\bf{B}$ is incorrect](1)</Checkpoint>

Assuming that there is only $\mathrm{AgCl}$ precipitate and no $\mathrm{Ag_2CrO_4}$ precipitate, then:

$\left[\mathrm{Ag}^{+}\right]=K_{\mathrm{sp}}{ }^{1 / 2}=\left(1.8 \times 10^{-10}\right)^{1 / 2}=1.34 \times 10^{-5}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$

<Checkpoint>[Assuming only one precipitate is formed, the calculated silver ion concentration is $1.34 \times 10^{-5}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$.](1)</Checkpoint>

The original concentration of the indicator solution is approximately:

$\left[\mathrm{K_2Cr_2 O_7}\right]=0.170 \mathrm{~mol} \cdot \mathrm{L}^{-1}$

The actual $\mathrm{CrO_4}^{2-}$ concentration at the end of the titration is:

$\left[\mathrm{CrO_4^{2-}}\right]=\frac{0.170 \times 2 \times 0.05 \times 6} {25.00+22.61+0.30}=2.1 \times 10^{-3}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$

<Checkpoint>[The actual $\mathrm{CrO_4^{2-}}$ concentration at the end of the titration is $2.1 \times 10^{-3}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$.](1)</Checkpoint>

To generate $\mathrm{Ag}_{2} \mathrm{Cr}\mathrm{O}_{4}$ precipitate, the concentration of $\mathrm{CrO_4^{2-}}$ must be at least:

$\left[\mathrm{CrO_4^{2-}}\right]=K_{\mathrm{sp}} /\left[\mathrm{Ag}^{+}\right]^{2}=6.6 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

<Checkpoint>[Therefore, no precipitate is formed.](1)</Checkpoint>

$\left[\mathrm{Ag}^{+}\right]=1.34 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

<Checkpoint>[$\bf{C}$、$\bf{D}$ are incorrect](1)</Checkpoint>

To make the endpoint error $0$, at the titration endpoint, we should have:

$\left[\mathrm{CrO_4^{2-}}\right]=6.6 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

Let $x \mathrm{~mL}$ of titrant be added, then $\left[\mathrm{CrO_4^{2-}}\right]=\frac{0.170 \times 2 \times x}{25.00+22.61+x}=6.6 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

<Checkpoint>[Solving the equation, we can get $x=0.94$](1)</Checkpoint>

The volume of indicator solution to be added is $0.94 \mathrm{mL}$.

<Checkpoint>[$\bf{E}$ is correct](1)</Checkpoint>

When using the $\bf{Mohr}$ method, $\mathrm{AgI}$ and $\mathrm{AgSCN}$ precipitates have strong adsorption, making the endpoint color change unclear and the error large.

<Checkpoint>[$\bf{F}$ is incorrect](1)</Checkpoint>
氯化钠含量的计算：

先求出滴定用去的$\mathrm{AgNO_3}$，进而得到$\mathrm{Cl^-}$的量。结合溶液稀释，反推至原始食盐样品中的$\mathrm{Cl}$的量，进而算出氯化钠含量。计算如下：

$$\omega=\frac{0.1000 \times(22.61-0.05) \times 10^{-3} \times(35.45+22.99) \times 10} {1.3567}=97.18 \%$$

<Checkpoint>[故氯化钠含量为$97.18 \%$。$\bf{B}$错误](1)</Checkpoint>

假设此时只有 $\mathrm{AgCl}$ 沉淀而没有 $\mathrm{Ag_2CrO_4}$ 沉淀, 则:

$\left[\mathrm{Ag}^{+}\right]=K_{\mathrm{sp}}{ }^{1 / 2}=\left(1.8 \times 10^{-10}\right)^{1 / 2}=1.34 \times 10^{-5}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$

<Checkpoint>[假设只有一种沉淀计算得到银离子浓度为$1.34 \times 10^{-5}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$。](1)</Checkpoint>

指示剂溶液原浓度约为: 

$\left[\mathrm{K_2Cr_2 O_7}\right]=0.170 \mathrm{~mol} \cdot \mathrm{L}^{-1}$

滴定结束时实际的 $\mathrm{CrO_4}^{2-}$ 浓度为:

$\left[\mathrm{CrO_4^{2-}}\right]=\frac{0.170 \times 2 \times 0.05 \times 6} {25.00+22.61+0.30}=2.1 \times 10^{-3}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$

<Checkpoint>[滴定结束时实际的 $\mathrm{CrO_4^{2-}}$ 浓度为$2.1 \times 10^{-3}\left(\mathrm{~mol} \cdot \mathrm{L}^{-1}\right)$。](1)</Checkpoint>

为生成 $\mathrm{Ag}_{2} \mathrm{Cr}\mathrm{O}_{4}$ 沉淀, 则 $\mathrm{CrO_4^{2-}}$ 的浓度至少为:

$\left[\mathrm{CrO_4^{2-}}\right]=K_{\mathrm{sp}} /\left[\mathrm{Ag}^{+}\right]^{2}=6.6 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

<Checkpoint>[因此无沉淀生成 。](1)</Checkpoint>

$\left[\mathrm{Ag}^{+}\right]=1.34 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

<Checkpoint>[$\bf{C}$、$\bf{D}$错误](1)</Checkpoint>

为使得终点误差为 $0$ , 滴定终点时应有: 

$\left[\mathrm{CrO_4^{2-}}\right]=6.6 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

设加入 $x \mathrm{~mL}$ 滴定剂, 则 $\left[\mathrm{CrO_4^{2-}}\right]=\frac{0.170 \times 2 \times x}{25.00+22.61+x}=6.6 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1}$

<Checkpoint>[解方程，可得 $x=0.94$](1)</Checkpoint>

应加入指示剂溶液的体积$0.94 \mathrm{mL}$。

<Checkpoint>[$\bf{E}$正确](1)</Checkpoint>

$\bf{Mohr}$ 法测定时，$\mathrm{AgI}$ 和 $\mathrm{AgSCN}$ 沉淀有强烈的吸附作用，使终点变色不明显，误差较大。

<Checkpoint>[$\bf{F}$错误](1)</Checkpoint>