$\bf M$ is a metallic element that can form a salt $\rm {\bf MX}_2\cdot nH_2O$ ($\rm n$ is an integer) with a monobasic organic acid $\rm H \bf X$. The following thermogravimetric analysis experiment was performed on the salt (all data are experimental values):

(1) $\rm {\bf MX}_2\cdot nH_2O$ is heated to 180 °C in nitrogen, with a weight loss of 14.2%, corresponding to the loss of all crystal water; heating is continued to 280 °C, with a further weight loss of 13.4% (Reaction 1); heating is continued to above 400 °C, and after cooling, the remaining solid is metal $\bf M$, with a mass of 54.8% of the original mass.

(2) $\rm {\bf MX}_2\cdot nH_2O$ is heated to above 400 °C in air, and after cooling, the remaining solid is oxide $\bf Y$, with a mass of 58.7% of the original mass.

Through calculation, deduce what metal element $\bf M$ is, and deduce what substances the salt $\rm {\bf MX}_2\cdot nH_2O$ and $\bf Y$ are. (Hint: The ratio of the number of metal atoms to oxygen atoms in $\bf Y$ is a simple integer ratio)

Which of the following statements are correct:
1. If $\bf M$ is in the a-th period and the b-th column from left to right, then (a+b)/ab>0.25
2. $\rm {\bf MX}_2$ is a strong electrolyte
3. $\bf Y$ is insoluble in $\rm NaOH$
4. The sum of the number of all atoms in the chemical formula of the solid product of Reaction 1 is less than or equal to 15
$\bf M$为金属元素，它能与一元有机酸$\rm H \bf X$形成盐$\rm {\bf MX}_2\cdot nH_2O$($\rm n$为整数)。对该盐进行了如下热重分析实验(数据均为实验值)：  

(1) $\rm {\bf MX}_2\cdot nH_2O$在氮气中加热至180 °C，失重14.2%，对应失去所有结晶水；继续加热至280 °C，再失重13.4%（反应1）；继续加热至400 °C以上，冷却后剩余固体为金属$\bf M$，质量为原质量的54.8%。  

(2) $\rm {\bf MX}_2\cdot nH_2O$在空气中加热至400 °C以上，冷却后剩余固体为氧化物$\bf Y$，质量为原质量的58.7%。  

通过计算，推出$\bf M$是何种金属元素，推出盐$\rm {\bf MX}_2\cdot nH_2O$和$\bf Y$各是何种物质。（提示：$\bf Y$中金属与氧原子的数目比为简单整数比）

下列说法正确的有：
1. 记$\bf M$在第a周期，从左至右第b列，则(a+b)/ab>0.25
2. $\rm {\bf MX}_2$是一种强电解质
3. $\bf Y$难溶于$\rm NaOH$
4. 反应1的固体产物化学式中所有原子数目之和小于等于15
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
Let the molar mass of $\bf M$ be $\rm x~g/mol$, then: $\rm 18.02n : x = 14.2 : 54.8$

When $\rm n=1$, $\rm x=69.5$ is $\rm Ga$ (oxidation state does not match, discard)

When $\rm n=2$, $\rm x=139$ is $\rm La$ (oxidation state does not match, discard) or $\rm Ba$

When $\rm n=3$, $\rm x=208.6$ is $\rm Bi$ (oxidation state does not match, discard) or $\rm Pb$

Combined with oxide data, let the chemical formula of oxide $\bf Y$ be $\rm MO_y$, then:

$\rm (x + 16.00y) : x = 58.7 : 54.8$

When $\bf M$ is $\rm Ba$, solving for $\rm y=0.60$, which is unreasonable, discard;

When $\bf M$ is $\rm Pb$, solving for $\rm y=0.92 \approx 1$, corresponding to $\rm PbO$;

Therefore, metal $\bf M$ is $\rm Pb$, $\bf Y$ is $\rm PbO$

<Checkpoint>[$\bf M$ is $\rm Pb$, $\bf Y$ is $\rm PbO$](2)</Checkpoint>

The molar mass of the acid radical ion is: $\rm (207.2 \div 0.548 - 207.2 - 3 \times 18.02)/2 = 58.4~(g/mol)$

Organic acids are often carboxylic acids, and the molecular weight of $\rm -COO^-$ is 44; therefore, 58.4 corresponds to acetate, and $\rm HX$ is acetic acid

$\rm {\bf MX}_2\cdot nH_2O$ is $\rm Pb(CH_3COO)_2\cdot 3H_2O$

<Checkpoint>[$\rm {\bf MX}_2\cdot nH_2O$ is $\rm Pb(CH_3COO)_2\cdot 3H_2O$](1)</Checkpoint>

a=6, b=14, (a+b)/ab=0.238, statement 1 is incorrect

<Checkpoint>[(a+b)/ab=0.238](0.5)</Checkpoint>

$\rm Pb(CH_3COO)_2$ is a weak electrolyte, statement 2 is incorrect

<Checkpoint>[$\rm Pb(CH_3COO)_2$ is a weak electrolyte](0.5)</Checkpoint>

$\rm PbO$ is amphoteric and soluble in $\rm NaOH$, statement 3 is incorrect

<Checkpoint>[$\rm PbO$ is amphoteric and soluble in $\rm NaOH$](0.5)</Checkpoint>

The molar mass of the solid product obtained after weight loss at 280 °C in nitrogen is (assuming it contains 1 $\rm Pb$):

$\rm (207.2 + 2 \times 59.04 + 3 \times 18.02) \times (1 - 0.142 - 0.134) = 274.6~(g/mol)$

This molar mass corresponds to 1 $\rm Pb$ atom, 1 acetate ion, and 0.5 oxygen atoms. Therefore, the solid product of reaction 1 is $\rm Pb_2O(CH_3COO)_2$, and the sum of the number of all atoms in the chemical formula is 17, statement 4 is incorrect

<Checkpoint>[The solid product of reaction 1 is $\rm Pb_2O(CH_3COO)_2$](0.5)</Checkpoint>

Choose A
设$\bf M$的摩尔质量为$\rm x~g/mol$，则：$\rm 18.02n : x = 14.2 : 54.8$

当$\rm n=1$时，$\rm x=69.5$为$\rm Ga$（氧化态不符，舍去）

当$\rm n=2$时，$\rm x=139$为$\rm La$（氧化态不符，舍去）或$\rm Ba$

当$\rm n=3$时，$\rm x=208.6$为$\rm Bi$（氧化态不符，舍去）或$\rm Pb$

结合氧化物数据，设氧化物$\bf Y$的化学式为$\rm MO_y$，则：

$\rm (x + 16.00y) : x = 58.7 : 54.8$

当$\bf M$为$\rm Ba$时，解得$\rm y=0.60$，不合理，舍去；

当$\bf M$为$\rm Pb$时，解得$\rm y=0.92 \approx 1$，对应$\rm PbO$；

因此，金属$\bf M$为$\rm Pb$，$\bf Y$为$\rm PbO$

<Checkpoint>[$\bf M$为$\rm Pb$，$\bf Y$为$\rm PbO$](2)</Checkpoint>

酸根离子摩尔质量为：$\rm (207.2 \div 0.548 - 207.2 - 3 \times 18.02)/2 = 58.4~(g/mol)$

有机酸常为羧酸，$\rm -COO^-$的分子量为44；因此58.4对应于乙酸根，$\rm HX$为乙酸

$\rm {\bf MX}_2\cdot nH_2O$为$\rm Pb(CH_3COO)_2\cdot 3H_2O$

<Checkpoint>[$\rm {\bf MX}_2\cdot nH_2O$为$\rm Pb(CH_3COO)_2\cdot 3H_2O$](1)</Checkpoint>

a=6，b=14，(a+b)/ab=0.238，说法1错误

<Checkpoint>[(a+b)/ab=0.238](0.5)</Checkpoint>

$\rm Pb(CH_3COO)_2$是弱电解质，说法2错误

<Checkpoint>[$\rm Pb(CH_3COO)_2$是弱电解质](0.5)</Checkpoint>

$\rm PbO$具有两性，可溶于$\rm NaOH$，说法3错误

<Checkpoint>[$\rm PbO$具有两性，可溶于$\rm NaOH$](0.5)</Checkpoint>

在氮气中280 °C下失重所得固体产物的摩尔质量为（假设含1个$\rm Pb$）：  

$\rm (207.2 + 2 \times 59.04 + 3 \times 18.02) \times (1 - 0.142 - 0.134) = 274.6~(g/mol)$  

该摩尔质量对应于1个$\rm Pb$原子，1个醋酸根离子和0.5个氧原子。因此，反应1的固体产物为$\rm Pb_2O(CH_3COO)_2$，化学式所有原子数目之和为17，说法4错误

<Checkpoint>[反应1的固体产物为$\rm Pb_2O(CH_3COO)_2$](0.5)</Checkpoint>

选A