A certain basic salt $\bf A$ of metal $\bf M$ is a bright green crystal. $\rm 2.524~g$ of this salt was placed in an inert atmosphere and heated to $\rm 148~^ \circ C $, resulting in a weight loss of 12.19%, corresponding to the loss of all crystal water; continued heating to $\rm 233~^ \circ C$, resulted in a weight loss of 9.26%; continued heating above $\rm 300~^ \circ C$ to constant weight, producing a gas with a density of $\rm 1.80~g/L~(25~^ \circ C , 101.3~kPa)$. The composition of the remaining solid is an oxide of $\bf M$, with a mass of $\rm 1.603~g$. Let a be the column number of $\bf M$ in the periodic table, and b be the total number of ions in $\bf A$, then a/b is approximately
金属$\bf M$的某种碱式盐$\bf A$为亮绿色晶体。将$\rm 2.524~g$该盐置于惰性气氛中，加热至$\rm 148~^ \circ C $，失重12.19%，对应于失去所有结晶水；继续加热至$\rm 233~^ \circ C$，失重9.26%；继续加热至$\rm 300~^ \circ C$以上至恒重，生成密度为$\rm 1.80~g/L~(25~^ \circ C , 101.3~kPa)$ 的气体。残余固体的成分为$\bf M$的氧化物，质量为$\rm 1.603~g$。记$\bf M$在周期表的第a列，$\bf A$中的离子数总和为b，则a/b约为
All other options are incorrect
0.41
0.59
0.69
0.77
0.91
0.98
1.23
1.46
1.65
1.88
2.01
2.47
1.06
1.33
0.33
其他选项均不正确
0.41
0.59
0.69
0.77
0.91
0.98
1.23
1.46
1.65
1.88
2.01
2.47
1.06
1.33
0.33
From $\rm pM=ρRT$, it can be known that the molecular weight of the gas lost in the third step is 44.0, which is $\rm CO_2$

<Checkpoint>[The gas lost in the third step is $\rm CO_2$](1)</Checkpoint>

$\bf A$ is a basic carbonate, and the first two steps can only be water loss. Assume that the first step loses crystal water and the second step loses hydroxyl water.

Moles of crystal water $\rm n_1(H_2O)=2.524~g×12.19\%/18.02~g·mol^{-1}=0.01707~mol$

<Checkpoint>[Crystal water is 0.01707 mol](0.5)</Checkpoint>

Moles of hydroxyl water $\rm n_2(H_2O)=2.524~g \times 9.26\%/18.02~g·mol^{-1}=0.01297~mol$

<Checkpoint>[Hydroxyl water is 0.01297 mol](0.5)</Checkpoint>

$\rm n(CO_2)=[2.524~g×(1-12.19\%-9.26\%)-1.603~g]/44.01~g·mol^{-1}=0.008625~mol$

<Checkpoint>[0.008625 mol $\rm CO_2$ is released in the third step](0.5)</Checkpoint>

Assuming that $\bf M$ in the remaining solid has a valence of +2, the number of moles of $\bf M \rm O$ (corresponding to $\rm CO_2$ and hydroxyl water) is: $\rm n( {\bf M}O)=0.01297~mol+0.008625~mol=0.02160~mol$

<Checkpoint>[$\bf M \rm O$ is 0.02160 mol](0.5)</Checkpoint>

The molar ratio of crystal water, hydroxyl water, $\rm CO_2$, and $\bf M$ is 4:3:2:5, so the chemical formula of $\bf A$ is $\bf M \rm _5(CO_3)_2(OH)_6·4H_2O$

<Checkpoint>[The chemical formula of $\bf A$ is $\bf M \rm _5(CO_3)_2(OH)_6·4H_2O$](1)</Checkpoint>

The molar mass of $\bf M \rm O$ is $\rm 1.603~g/0.02160~mol=74.21~g·mol^{-1}$, $\bf M$ is $\rm Ni$

<Checkpoint>[$\bf M$ is $\rm Ni$](0.5)</Checkpoint>

If $\bf M$ has other valences, there are no reasonable results.

a=10, b=13, a/b=0.77, select E

<Checkpoint>[a/b=0.77](0.5)</Checkpoint>
由$\rm pM=ρRT$可知，第三步失去的气体分子量为44.0，为$\rm CO_2$

<Checkpoint>[第三步失去的气体为$\rm CO_2$](1)</Checkpoint>

$\bf A$为碱式碳酸盐，前两步只能是失水。假设第一步失去结晶水，第二步失去羟基水。

结晶水摩尔数$\rm n_1(H_2O)=2.524~g×12.19\%/18.02~g·mol^{-1}=0.01707~mol$

<Checkpoint>[结晶水为0.01707 mol](0.5)</Checkpoint>

羟基水摩尔数$\rm n_2(H_2O)=2.524~g \times 9.26\%/18.02~g·mol^{-1}=0.01297~mol$

<Checkpoint>[羟基水为0.01297 mol](0.5)</Checkpoint>

$\rm n(CO_2)=[2.524~g×(1-12.19\%-9.26\%)-1.603~g]/44.01~g·mol^{-1}=0.008625~mol$

<Checkpoint>[第三步放出0.008625 mol $\rm CO_2$](0.5)</Checkpoint>

假设剩下的固体中$\bf M$为+2价，则$\bf M \rm O$摩尔数为（对应$\rm CO_2$和羟基水）：$\rm n( {\bf M}O)=0.01297~mol+0.008625~mol=0.02160~mol$

<Checkpoint>[$\bf M \rm O$为0.02160 mol](0.5)</Checkpoint>

结晶水、羟基水、$\rm CO_2$和$\bf M$的摩尔比为4:3:2:5，故$\bf A$化学式为$\bf M \rm _5(CO_3)_2(OH)_6·4H_2O$

<Checkpoint>[$\bf A$化学式为$\bf M \rm _5(CO_3)_2(OH)_6·4H_2O$](1)</Checkpoint>

$\bf M \rm O$摩尔质量为$\rm 1.603~g/0.02160~mol=74.21~g·mol^{-1}$，$\bf M$为$\rm Ni$

<Checkpoint>[$\bf M$为$\rm Ni$](0.5)</Checkpoint>

若$\bf M$为其他价态，均没有合理的结果

a=10, b=13, a/b=0.77，选E

<Checkpoint>[a/b=0.77](0.5)</Checkpoint>