The laboratory uses the following method to determine the content of mercury sulfide $(HgS)$ in cinnabar samples:

(1) Weigh $3.2643 g$ of dried primary standard sodium chloride, dissolve it in water, and dilute to a $500.0 mL$ volumetric flask.

(2) Transfer $25.00mL$ of sodium chloride solution, add $50.00mL$ of silver nitrate solution, $5mL$ of nitric acid, and 10 drops of indicator $\mathbf{A}$, and titrate with $NH_4SCN$ solution to the endpoint, consuming $23.97mL$.

(3) Transfer $25.00mL$ of silver nitrate solution, add $5mL$ of nitric acid, and 10 drops of indicator $\mathbf{A}$, and titrate with $NH_4SCN$ solution to the endpoint, consuming $26.26mL$.

(4) Weigh another $0.2846g$ of the cinnabar sample into another test tube, add $10mL$ of sulfuric acid and $1.5g$ of potassium nitrate, and heat to dissolve, producing a reddish-brown gas and a pale yellow precipitate; after cooling, add $10g\cdot L^{-1}$ potassium permanganate solution dropwise until pink, the precipitate disappears, and then boil the solution until colorless, and no precipitate is formed in this step.

(5) Add ten drops of indicator $\mathbf{A}$, and titrate with the above ammonium thiocyanate solution to the endpoint, consuming $22.83mL$.

Regarding the above experimental process, there are the following statements:

(1) The concentration of the $NaCl$ solution is $0.112mol\cdot L^{-1}$

(2) The concentration of the $NH_4SCN$ solution is $0.1182mol\cdot L^{-1}$

(3) The concentration of the $AgNO_3$ solution is $0.1125mol\cdot L^{-1}$

(4) In the chemical reaction equation for dissolving the cinnabar sample, if the equation is balanced according to the simplest reaction ratio, the reaction coefficient of $H_2SO_4$ is 3

(5) When adding potassium permanganate solution dropwise, if the ion equation is balanced according to the simplest reaction ratio, the coefficient of $H_2O$ in the equation is 4

(6) $\mathbf{A}$ in the question can be ferric chloride

(7) The content of mercury sulfide in the cinnabar sample is $91.27\%$

Among the above statements, the number of correct ones is (pay attention to significant figures, if the number of significant figures is too small, it should be treated as incorrect; for options involving calculations, if the error is less than $1\%$, it is considered correct)
实验室用如下方法测定朱砂样品中硫化汞$(HgS)$的含量：

 （1）称取干燥好的基准氯化钠$3.2643 g$，溶于水，稀释至$500.0 mL$容量瓶中

（2）移取$25.00mL$氯化钠溶液，加入$50.00mL$硝酸银溶液，$5mL$硝酸和10滴指示剂$\mathbf{A}$，用$NH_4SCN$溶液滴定至终点，消耗$23.97mL$。

（3）移取$25.00mL$硝酸银溶液，加入$5mL$硝酸和10滴指示剂$\mathbf{A}$，用$NH_4SCN$溶液滴定至终点，消耗$26.26mL$。

（4）再称取朱砂样品$0.2846g$于另一试管，加入$10mL$硫酸和$1.5g$硝酸钾加热溶解，生成一种红棕色气体与淡黄色沉淀；待冷却后逐滴加入$10g\cdot L^{-1}$高锰酸钾溶液至粉红色，沉淀消失，再煮沸溶液至无色，该步骤无沉淀生成。

（5）加指示剂$\mathbf{A}$十滴，并用上述硫氰酸铵溶液滴定至终点，消耗$22.83mL$。

对于上述实验过程，有以下几种说法：

（1）$NaCl$溶液的浓度为$0.112mol\cdot L^{-1}$

（2）$NH_4SCN$溶液的浓度为$0.1182mol\cdot L^{-1}$

（3）$AgNO_3$溶液的浓度为$0.1125mol\cdot L^{-1}$

（4）在溶解朱砂样品的化学反应方程式中，按最简反应比配平方程式，则$H_2SO_4$的反应系数为3

（5）逐滴加入高锰酸钾溶液时，按最简反应比配平其离子方程式，方程式中$H_2O$的系数为4

（6）题目中$\mathbf{A}$可以是三氯化铁

（7）该朱砂样品中硫化汞的含量为$91.27\%$

以上说法中，正确的个数为（注意有效数字，有效数字位数过少应当错误处理，涉及计算的选项，误差小于$1\%$即认为正确）
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
Statement (1):

Relative molecular mass of sodium chloride: $M_{NaCl}=22.99+35.45=58.44g\cdot mol^{-1}$

Concentration of sodium chloride standard solution: $c_{NaCl}=\frac{m_{NaCl}}{M_{NaCl}\cdot V_1}=\frac{3.2643g}{58.44g\cdot mol^{-1}\times 500.0mL}=0.1117 mol\cdot L^{-1}$

According to the significant figures principle of analytical chemistry, 4 to 5 significant figures should be retained. Statement (1) only retains 3 significant figures, so **statement (1) is incorrect**

<Checkpoint>[The concentration of sodium chloride solution is $0.1117 mol/L$](1)</Checkpoint>

<Checkpoint>[4-5 significant figures should be retained](2)</Checkpoint>

Statement (2) (3):

According to step (2) and step (3), two equations can be listed:

$c_{NH_4SCN}\cdot V_4+c_{NaCl}\cdot V_2=c_{AgNO_3}\cdot V_总$

$c_{AgNO_3}\cdot V_3'=c_{NH_4SCN}\cdot V_4'$

Assume the concentration of $NH_4SCN$ is $y$, the concentration of $AgNO_3$ is $x$, substitute the data to get

$y*23.97 mL+0.1117mol/L\times 25.00mL=x*50.00mL$

$x*25.00mL=y*26.26mL$

Solving the two equations simultaneously:
$x=0.1027 mol/L, y=0.09781 mol/L$

Therefore, **statement (2) and statement (3) are incorrect**

<Checkpoint>[The concentration of silver nitrate solution is $0.1027mol/L$](1)</Checkpoint>

<Checkpoint>[The concentration of ammonium thiocyanate solution is $0.09781mol/L$](1)</Checkpoint>

Statement (4):

According to "a light yellow precipitate is formed", it can be judged that the oxidation product is $S$. From the reddish-brown gas, it can be judged that the reduction product is $NO_2$. The equation can be written as

$HgS +2H_2SO_4 + 2KNO_3 = 2NO_2 + HgSO_4 + 2H_2O + S + K_2SO_4$

The coefficient of $H_2SO_4$ is 2, **statement (4) is incorrect**

<Checkpoint>[$HgS +2H_2SO_4 + 2KNO_3 = 2NO_2 + HgSO_4 + 2H_2O + S + K_2SO_4$](1)</Checkpoint>

Statement (5):

In the process of adding $KMnO_4$ solution dropwise, $MnO_4^-$ oxidizes $S$ element to $SO_4^{2-}$, and itself is reduced to $Mn^{2-}$. Write the ionic equation under acidic conditions:

$5S + 6MnO_4^- + 8H^+ = 6Mn^{2+} + 5SO_4^{2-} + 4H_2O$

Therefore, the coefficient of $H_2O$ is 4, **statement (5) is correct**

<Checkpoint>[$5S + 6MnO_4^- + 8H^+ = 6Mn^{2+} + 5SO_4^{2-} + 4H_2O$](1)</Checkpoint>

Statement (6):

The function of indicator $\mathbf{A}$ is to identify excess $SCN^-$ to prove that $Ag^+$ has completely reacted. This titration is a variant of Mohr's method or Volhard's method, where Volhard's method uses $Fe^{3+}$ as an indicator to form a blood-red $[Fe(SCN)]^{2+}$ complex with excess $SCN^-$, indicating the titration endpoint. **Statement (6) is correct**

<Checkpoint>[$Fe^{3+}$ is used as an indicator to generate blood-red $[Fe(SCN)]^{2+}$](1)</Checkpoint>

Statement (7):

The titration reaction equation is:

$Hg^{2+} + 2SCN^- = Hg(SCN)_2\downarrow$

$n(Hg)=n(SCN^-)/2=\frac{1}{2}\cdot c_{NH_4SCN}V_6=\frac{1}{2}\times 0.09781mol/L\times 22.83mL=1.116\times 10^{-3}mol$

The relative molecular mass of mercury sulfide is

$M(HgS)=200.6+32.07=232.67g/mol$

$\omega(HgS)=\frac{M(HgS)n(Hg)}{m}=\frac{232.67g/mol\times 1.116\times 10^{-3}mol}{0.2846g}=91.27\%$

Therefore, **statement (7) is correct**

<Checkpoint>[$Hg^{2+} + 2SCN^- = Hg(SCN)_2\downarrow$](1)</Checkpoint>

<Checkpoint>[The mass fraction of mercury sulfide in cinnabar is $91.27\%$](1)</Checkpoint>

Statements (5), (6), and (7) are all correct, so the answer to this question is $D$
说法（1）：

氯化钠的相对分子质量：$M_{NaCl}=22.99+35.45=58.44g\cdot mol^{-1}$

氯化钠标准溶液的浓度：$c_{NaCl}=\frac{m_{NaCl}}{M_{NaCl}\cdot V_1}=\frac{3.2643g}{58.44g\cdot mol^{-1}\times 500.0mL}=0.1117 mol\cdot L^{-1}$

根据分析化学的有效数字原则，应保留4到5位有效数字，说法（1）仅保留了3位有效数字，因此**说法（1）错误**

<Checkpoint>[氯化钠溶液的浓度为$0.1117 mol/L$](1)</Checkpoint>

<Checkpoint>[应保留4-5位有效数字](2)</Checkpoint>

说法（2）（3）：

根据步骤（2）和步骤（3）可以列出两个方程：

$c_{NH_4SCN}\cdot V_4+c_{NaCl}\cdot V_2=c_{AgNO_3}\cdot V_总$

$c_{AgNO_3}\cdot V_3'=c_{NH_4SCN}\cdot V_4'$

假设$NH_4SCN$的浓度为$y$，$AgNO_3$的浓度为$x$，带入数据得

$y*23.97 mL+0.1117mol/L\times 25.00mL=x*50.00mL$

$x*25.00mL=y*26.26mL$

联立两方程解得：
$x=0.1027 mol/L, y=0.09781 mol/L$

因此**说法（2），说法（3）错误**

<Checkpoint>[硝酸银溶液的浓度为$0.1027mol/L$](1)</Checkpoint>

<Checkpoint>[硫氰酸铵溶液的浓度为$0.09781mol/L$](1)</Checkpoint>

说法（4）：

根据“生成淡黄色沉淀”可以判断氧化产物为$S$，由红棕色气体可判断还原产物为$NO_2$，可写出方程式

$HgS +2H_2SO_4 + 2KNO_3 = 2NO_2 + HgSO_4 + 2H_2O + S + K_2SO_4$

$H_2SO_4$的系数为2，**说法（4）错误**

<Checkpoint>[$HgS +2H_2SO_4 + 2KNO_3 = 2NO_2 + HgSO_4 + 2H_2O + S + K_2SO_4$](1)</Checkpoint>

说法（5）：

逐滴加入$KMnO_4$溶液的过程中，$MnO_4^-$将$S$单质氧化为$SO_4^{2-}$，自身还原为$Mn^{2-}$，写出酸性条件下的离子方程式：

$5S + 6MnO_4^- + 8H^+ = 6Mn^{2+} + 5SO_4^{2-} + 4H_2O$

因此$H_2O$的系数为4，**说法（5）正确**

<Checkpoint>[$5S + 6MnO_4^- + 8H^+ = 6Mn^{2+} + 5SO_4^{2-} + 4H_2O$](1)</Checkpoint>

说法（6）：

指示剂$\mathbf{A}$的作用是鉴别过量的$SCN^-$以证明$Ag^+$完全反应，该滴定是莫尔法或佛尔哈德法的一种变体，其中佛尔哈德法使用$Fe^{3+}$作指示剂，与过量的$SCN^-$生成血红色的$[Fe(SCN)]^{2+}$配合物，指示滴定终点。**说法（6）正确**

<Checkpoint>[$Fe^{3+}$作指示剂，生成血红色的$[Fe(SCN)]^{2+}$](1)</Checkpoint>

说法（7）：

滴定反应方程式为：

$Hg^{2+} + 2SCN^- = Hg(SCN)_2\downarrow$

$n(Hg)=n(SCN^-)/2=\frac{1}{2}\cdot c_{NH_4SCN}V_6=\frac{1}{2}\times 0.09781mol/L\times 22.83mL=1.116\times 10^{-3}mol$

硫化汞的相对分子质量为

$M(HgS)=200.6+32.07=232.67g/mol$

$\omega(HgS)=\frac{M(HgS)n(Hg)}{m}=\frac{232.67g/mol\times 1.116\times 10^{-3}mol}{0.2846g}=91.27\%$

因此**说法（7）正确**

<Checkpoint>[$Hg^{2+} + 2SCN^- = Hg(SCN)_2\downarrow$](1)</Checkpoint>

<Checkpoint>[朱砂中硫化汞的质量分数为$91.27\%$](1)</Checkpoint>

说法（5）（6）（7）均正确，本题选$D$