Sodium azide is a commonly used reagent in organic chemistry. In the laboratory, the purity of sodium azide is often determined by titration analysis, and the specific steps are as follows:  

1. Accurately weigh $0.4620 \mathrm{~g}$ of dried arsenic trioxide into a $250 \mathrm{~mL}$ beaker, add $100 \mathrm{~mL}$ of $100 \mathrm{~g} \cdot \mathrm{L}^{-1}$ sodium hydroxide solution to dissolve it, then transfer to a $1 \mathrm{~L}$ volumetric flask. Add $250 \mathrm{~mL}$ of water and $250 \mathrm{~mL}$ of $3 \mathrm{~mol} \cdot \mathrm{L}^{-1}$ sulfuric acid, and dilute to the mark.  

2. Precisely pipette $50.00 \mathrm{~mL}$ of arsenic trioxide standard solution into a $500 \mathrm{~mL}$ conical flask, add 3 drops of $2.5 \mathrm{~g} \cdot \mathrm{L}^{-1}$ osmium tetroxide solution, then add $25.00 \mathrm{~mL}$ of the ceric ammonium nitrate solution to be standardized. Dilute with water to $150 \mathrm{~mL}$, add $30 \mathrm{~mL}$ of $3 \mathrm{~mol} \cdot \mathrm{L}^{-1}$ sulfuric acid and 2 drops of ferroin indicator, and titrate with the ferrous ammonium sulfate solution to be standardized until the endpoint is reached. The titration consumes $\mathrm{39.90 ~mL}$ of ferrous ammonium sulfate standard solution.  

3. Using all the reagents mentioned above except arsenic trioxide, the titration consumes $\mathrm{49.18 ~mL}$ of ferrous ammonium sulfate standard solution.  

4. Accurately weigh $\mathrm{2.7233 ~g}$ of sodium azide sample, dissolve it in water, and dilute to $\mathrm{250 ~mL}$. Pipette $\mathrm{20.00 ~mL}$ of this solution into a $\mathrm{500 ~mL}$ conical flask, add $\mathrm{40.00 ~mL}$ of ceric ammonium nitrate standard solution, and allow the reaction to proceed in a cold water bath. After the reaction is complete, dilute to $\mathrm{150 ~mL}$, add $\mathrm{30 ~mL\ 3 ~mol\cdot L^{-1}}$ sulfuric acid and 2 drops of ferroin indicator, then titrate with ferrous ammonium sulfate standard solution until the endpoint is reached. The titration consumes $\mathrm{15.32 ~mL}$ of ferrous ammonium sulfate standard solution.  

Regarding this analytical process, the following statements are made:  

A. Dried arsenic trioxide can be directly used to prepare the standard test solution.  

B. The osmium tetroxide added undergoes a change in oxidation state during the analysis.  

C. The ionic equation for the chemical reaction occurring after adding ceric ammonium nitrate in step 2, when balanced with the simplest integer coefficients, has a sum of product coefficients equal to 5.  

D. The concentration of the prepared arsenic trioxide solution is $\mathrm{0.002335 ~mol/L}$.  

E. The concentration of the ceric ammonium nitrate solution is $\mathrm{0.09900 ~mol/L}$.  

F. The concentration of ferrous ammonium sulfate is $\mathrm{0.04032 ~mol/L}$.  

G. The content of sodium azide is $\mathrm{95.16 ~\%}$.  

Among these statements, the number of correct ones is (for calculation-related options, an error less than $\mathrm{1 ~\%}$ is considered correct).
叠氮化钠是有机化学中常用的反应试剂。实验室中常采用滴定分析的方法测定叠氮化钠的纯度，其具体步骤如下:

1. 准确称取 $0.4620 \mathrm{~g}$ 干燥的三氧化二砷于 $250 \mathrm{~mL}$ 烧杯中,加入 $100 \mathrm{~mL}\ 100 \mathrm{~g} \cdot \mathrm{L}^{-1}$ 氢氧化钠溶液溶解并转移至 $1 \mathrm{~L}$ 容量瓶中,加入 $250 \mathrm{~mL}$ 水和 $250 \mathrm{~mL}\ 3 \mathrm{~mol} \cdot \mathrm{L}^{-1}$ 硫酸,定容。

2. 准确移取 $50.00 \mathrm{~mL}$ 三氧化二砷标准溶液于 $500 \mathrm{~mL}$ 锥形瓶中,加入3滴 $2.5 \mathrm{~g} \cdot \mathrm{L}^{-1}$ 的四氧化锇溶液,再加入 $25.00 \mathrm{~mL}$ 待标定的硝酸铈铵溶液,加水稀释至 $150 \mathrm{~mL}$,加入 $30 \mathrm{~mL}\  3\mathrm{~mol} \cdot \mathrm{L}^{-1}$ 硫酸，2滴邻二氮菲亚铁指示剂,用待标定的硫酸亚铁铵溶液滴定至终点,消耗$\mathrm{39.90 ~mL}$硫酸亚铁铵标准溶液。

3. 用上述除三氧化二砷外的所有试剂,消耗硫酸亚铁铵标准溶液 $49.18 \mathrm{~mL}$。

4. 准确称取$2.7233 \mathrm{~g}$叠氮化钠样品,溶于水后定容至$\mathrm{250 ~mL}$。用移液管移取$\mathrm{20.00 ~mL}$
上述溶液于$\mathrm{500 ~mL}$ 锥形瓶中,加入$\mathrm{40.00 ~mL}$硝酸铈铵标准溶液,在冷水浴下反应。反应结束后,稀释至$\mathrm{150 ~mL}$,加入$\mathrm{30 ~mL\ 3 ~mol\cdot L^{-1}}$硫酸, 2滴邻二氮菲亚铁指示剂,用硫酸亚铁
铵标准溶液滴定至终点,消耗硫酸亚铁铵标准溶液$\mathrm{15.32 ~mL}$。

关于该分析过程，有下列说法：

A. 干燥的三氧化二砷可以直接用于配置基准测试溶液。

B. 加入的四氧化锇在分析过程中有化合价的变化。

C. 过程2中加入硝酸铈铵后发生的化学反应的离子方程式，在系数为最简整数比配平后，生成物的系数和为5。

D. 配置的三氧化二砷溶液的浓度为$\mathrm{0.002335 ~mol/L}$

E. 硝酸铈铵溶液的浓度为$\mathrm{0.09900 ~mol/L}$

F. 硫酸亚铁铵的浓度为$\mathrm{0.04032 ~mol/L}$

G. 叠氮化钠含量为$\mathrm{95.16 ~\%}$

以上说法中，正确的个数为（涉及计算的选项，计算误差小于$\mathrm{1 ~\%}$即算正确。）
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Arsenic trioxide is a primary standard substance with strong stability. After drying, it can be used to directly prepare standard solutions, so statement A is correct.

<Checkpoint>[Arsenic trioxide is a primary standard substance, so statement A is correct](1)</Checkpoint>

The added osmium tetroxide acts as a catalyst in the analysis process, catalyzing the oxidation reaction. Osmium tetroxide oxidizes arsenic trioxide and is reduced to the hexavalent state itself. Ammonium cerium(IV) nitrate then oxidizes hexavalent osmium to osmium tetroxide, forming a catalytic cycle with a change in oxidation state, so statement B is correct.

<Checkpoint>[Osmium tetroxide acts as a catalyst to form an oxidation cycle, so statement B is correct](1)</Checkpoint>

In the analysis process, the main purpose of process 1 is to prepare a standard solution of arsenic trioxide, which dissolves in sodium hydroxide to produce arsenite ions $\mathrm{AsO_3^{3-}}$, and then is acidified to arsenous acid $\mathrm{H_3AsO_3}$. The concentration of the arsenic trioxide standard solution can be calculated as follows:

$c(\mathrm{As}) = n/V = 0.4620/197.84/1 = 0.002335 ~\mathrm{mol/L}$, so statement D is correct.

<Checkpoint>[$c(\mathrm{As}) = 0.002335 ~\mathrm{mol/L}$, so statement D is correct.](1)</Checkpoint>

Processes 2 and 3 are used to simultaneously standardize ferrous ammonium sulfate and ammonium cerium(IV) nitrate solutions. Ammonium cerium(IV) nitrate is used to oxidize arsenic trioxide, and the remaining ammonium cerium(IV) nitrate is back-titrated with ferrous ammonium sulfate solution; at the same time, a blank experiment without arsenic trioxide is used to determine the concentration of ferrous ammonium sulfate. The reactions that occur during the process are:

$\mathrm{H_3AsO_3 + 2Ce^{4+} + H_2O = 2Ce^{3+} + H_3AsO_4 + 2H^+}$

<Checkpoint>[$\mathrm{H_3AsO_3 + 2Ce^{4+} + H_2O = 2Ce^{3+} + H_3AsO_4 + 2H^+}$](1)</Checkpoint>

The sum of the coefficients of the products of this reaction is 5, so statement C is correct.

<Checkpoint>[The sum of the coefficients of the products of this reaction is 5, so statement C is correct.](0.5)</Checkpoint>

The back-titration equation is

$\mathrm{Ce^{4+} + Fe^{2+} = Ce^{3+} + Fe^{3+}}$

The difference in the amount of ferrous ammonium sulfate used in processes 2 and 3 corresponds to the amount of ammonium cerium(IV) nitrate that oxidized the arsenic trioxide; the amount of arsenic trioxide added is

<Checkpoint>[The difference in the amount of ferrous ammonium sulfate used corresponds to the amount of ammonium cerium(IV) nitrate that oxidized the arsenic trioxide](1)</Checkpoint>

$n = cV = 0.002335\times0.0500 = 0.00011675\mathrm{~mol}$

According to the redox stoichiometric relationship, $\mathrm{c(Fe^{2+})-c(Ce^{4+})-2c(H_3AsO_3)-4c(As_2O_3)}$

The concentration of ferrous ammonium sulfate is obtained as

$c(\mathrm{Fe})=n/V=\frac{4\times0.00011675}{(49.18-39.90)\times0.001}=0.05032 \mathrm{~mol/L}$

<Checkpoint>[$c(\mathrm{Fe})=0.05032 \mathrm{~mol/L}$, so statement F is incorrect](1)</Checkpoint>

Therefore, statement F is incorrect.

Through the blank experiment, all the ferrous ammonium sulfate in process 3 is used to reduce ammonium cerium(IV) nitrate, from which the concentration of ammonium cerium(IV) nitrate can be deduced:

$c(\mathrm{Ce})=c(\mathrm{Fe})\frac{V(\mathrm{Fe})}{V(\mathrm{Ce})}=0.05032\times\frac{49.18}{25.00}=0.09895 \mathrm{~mol/L}$

Therefore, statement E is correct.

<Checkpoint>[$c(\mathrm{Ce})=0.09895 \mathrm{~mol/L}$ Statement E is correct.](1)</Checkpoint>

Process 4 uses ammonium cerium(IV) nitrate to titrate sodium azide, and then ferrous ammonium sulfate is used to back-titrate the ammonium cerium(IV) nitrate. The equation for the first step of oxidizing sodium azide is

$\mathrm{2N_3^- + 2Ce^{4+} = 2Ce^{3+} + 3N_2}$

<Checkpoint>[$\mathrm{2N_3^- + 2Ce^{4+} = 2Ce^{3+} + 3N_2}$](1)</Checkpoint>

According to the reaction stoichiometry, one molecule of sodium azide corresponds to one molecule of ammonium cerium(IV) nitrate, so the amount of sodium azide is calculated as

$n(\mathrm{N_3^-})=n(\mathrm{Ce_{react}})=n(\mathrm{Ce_{total}})-n(\mathrm{Fe})=0.09895\times0.04-0.05030\times0.01532=0.0031874 \mathrm{~mol}$

$\omega(\mathrm{N_3^-})=m(\mathrm{N_3^-})/m(total)=0.0031874\times65.02\times(250/20)/2.7233=0.9513$

<Checkpoint>[$\omega(\mathrm{N_3^-})=0.9513$, so statement G is correct](1)</Checkpoint>

That is, the content is $95.13~\%$, statement G is correct.

In summary, there are six correct statements, so choose option G.
三氧化二砷为基准物质，稳定性强，其干燥后可用于直接配置标准溶液，说法A正确。

<Checkpoint>[三氧化二砷为基准物质，说法A正确](1)</Checkpoint>

加入的四氧化锇在分析过程中作为催化剂催化氧化反应，四氧化锇氧化三氧化二砷，自身被还原到六价，硝酸铈铵再氧化六价锇至四氧化锇形成催化循环，有化合价变化，说法B正确。

<Checkpoint>[四氧化锇作为催化剂形成氧化循环，说法B正确](1)</Checkpoint>

分析过程中，过程1主要目的为配置三氧化二砷标准溶液，其溶于氢氧化钠产生亚砷酸根离子$\mathrm{AsO_3^{3-}}$，再被酸化为亚砷酸$\mathrm{H_3AsO_3}$。由此可计算三氧化二砷标准溶液的浓度：

$c(\mathrm{As}) = n/V = 0.4620/197.84/1 = 0.002335 ~\mathrm{mol/L}$，说法D正确。

<Checkpoint>[$c(\mathrm{As}) = 0.002335 ~\mathrm{mol/L}$，说法D正确。](1)</Checkpoint>

过程2和3用于同时标定硫酸亚铁铵和硝酸铈铵两种溶液。利用硝酸铈铵氧化三氧化二砷，剩余的硝酸铈铵用硫酸亚铁铵溶液反滴定；同时利用不加三氧化二砷的空白实验测定硫酸亚铁铵的浓度。过程中发生的反应为：

$\mathrm{H_3AsO_3 + 2Ce^{4+} + H_2O = 2Ce^{3+} + H_3AsO_4 + 2H^+}$

<Checkpoint>[$\mathrm{H_3AsO_3 + 2Ce^{4+} + H_2O = 2Ce^{3+} + H_3AsO_4 + 2H^+}$](1)</Checkpoint>

该反应生成物系数和为5，说法C正确。

<Checkpoint>[该反应生成物系数和为5，说法C正确。](0.5)</Checkpoint>

返滴定方程式为

$\mathrm{Ce^{4+} + Fe^{2+} = Ce^{3+} + Fe^{3+}}$

过程2，3两次硫酸亚铁铵的使用量差值对应了硝酸铈铵氧化三氧化二砷的量；加入的三氧化二砷量为

<Checkpoint>[硫酸亚铁铵的使用量差值对应了硝酸铈铵氧化三氧化二砷的量](1)</Checkpoint>

$n = cV = 0.002335\times0.0500 = 0.00011675\mathrm{~mol}$

根据氧化还原计量关系，$\mathrm{c(Fe^{2+})-c(Ce^{4+})-2c(H_3AsO_3)-4c(As_2O_3)}$

得到硫酸亚铁铵的浓度

$c(\mathrm{Fe})=n/V=\frac{4\times0.00011675}{(49.18-39.90)\times0.001}=0.05032 \mathrm{~mol/L}$

<Checkpoint>[$c(\mathrm{Fe})=0.05032 \mathrm{~mol/L}$,说法F错误](1)</Checkpoint>

从而说法F错误。

通过空白实验，过程3的硫酸亚铁铵全用于还原硝酸铈铵，从而可推的硝酸铈铵的浓度：

$c(\mathrm{Ce})=c(\mathrm{Fe})\frac{V(\mathrm{Fe})}{V(\mathrm{Ce})}=0.05032\times\frac{49.18}{25.00}=0.09895 \mathrm{~mol/L}$

故说法E正确。

<Checkpoint>[$c(\mathrm{Ce})=0.09895 \mathrm{~mol/L}$  说法E正确。](1)</Checkpoint>

过程4利用硝酸铈铵滴定叠氮化钠，再采用硫酸亚铁铵返滴定硝酸铈铵。第一步氧化叠氮化钠的方程式为

$\mathrm{2N_3^- + 2Ce^{4+} = 2Ce^{3+} + 3N_2}$

<Checkpoint>[$\mathrm{2N_3^- + 2Ce^{4+} = 2Ce^{3+} + 3N_2}$](1)</Checkpoint>

根据反应计量关系，一份子叠氮化钠对应一分子硝酸铈铵，从而叠氮化钠的量计算为

$n(\mathrm{N_3^-})=n(\mathrm{Ce_{react}})=n(\mathrm{Ce_{total}})-n(\mathrm{Fe})=0.09895\times0.04-0.05030\times0.01532=0.0031874 \mathrm{~mol}$

$\omega(\mathrm{N_3^-})=m(\mathrm{N_3^-})/m(total)=0.0031874\times65.02\times(250/20)/2.7233=0.9513$

<Checkpoint>[$\omega(\mathrm{N_3^-})=0.9513$，说法G正确](1)</Checkpoint>

即含量为$95.13~\%$，G说法正确。

综上共有六个说法正确，选择选项G。