At $273 \mathrm{K}$, a mass $m$ of NaCl solid (whose specific heat can be neglected) is sprinkled into a mixture containing $100 \ \mathrm{g}$ of ice and $100 \ \mathrm{g}$ of water. Assuming the formed solution is an ideal mixture, neglecting the thermal effect of NaCl dissolution and heat exchange between the mixture and the surroundings, calculate the required mass $m$ of NaCl and the mass of ice at this point to ensure the final temperature of the ice-water mixture is $268 \mathrm{K}$.

Data that may be used for this problem:  
The specific heat capacity of water is $4.18 ~ \mathrm{J/(g \cdot K)}$,  
The specific heat capacity of ice is $2.10 ~ \mathrm{J/(g \cdot K)}$,  
The enthalpy of fusion of ice at $298 \mathrm{K}$ is $6008 ~ \mathrm{J/mol}$.  

Hint: When the amount of NaCl added is large, using approximate formulas may lead to errors.
$2 7 3 \mathrm { K }$ 下，向含 $1 0 0 \ \mathrm { g }$ 冰与 $1 0 0 \ \mathrm { g }$ 水的混合物中撒入质量为 $m$ 的NaCl固体(其比热可忽略不计)，假定形成的溶液是理想混合物，忽略NaCl溶解的热效应及混合物与外界的热交换，计算要使得最终所得冰水混合物的温度为268K所需要的NaCl的质量 $m$ 及此时冰的质量。

本题可能用到的数据：水的比热容为 $4 . 1 8 ~ \mathrm { J / ( g \cdot K ) }$ ，冰的比热容为 $2 . 1 0 \mathrm { J / ( g \cdot K ) }$ ， $2 9 8 \mathrm { K }$ 下冰的熔化焓为 $6 0 0 8 ~ \mathrm { J / m o l } _ { \circ }$

提示：NaCl的加入量较大时，使用近似式计算会导致误差。
The mass of NaCl required is 7.78 g, and the mass of ice is 88.5 g.
The required mass of NaCl is 17.3 g, and the mass of ice is 90.3 g.
The required mass of NaCl is 15.5 g, and the mass of ice is 88.5 g.
The required mass of NaCl is 9.78 g, and the mass of ice is 68.5 g.
The required mass of NaCl is 8.61 g, and the mass of ice is 90.3 g.
None of the above results are correct (with an error greater than 5%).
所需NaCl的质量为7.78 g，冰的质量为88.5 g
所需NaCl的质量为17.3 g，冰的质量为90.3 g
所需NaCl的质量为15.5 g，冰的质量为88.5 g
所需NaCl的质量为9.78 g，冰的质量为68.5 g
所需NaCl的质量为8.61 g，冰的质量为90.3 g
以上结果都不正确（有大于5%的误差）
In this problem, due to the large mass of NaCl, directly applying the freezing point depression law would cause errors!  

$\mathrm { H } _ { 2 } \mathrm { O } ( l ) = \mathrm { H } _ { 2 } \mathrm { O } ( s )$  

$\Delta _ { \mathrm { r } } C _ { \mathrm { m } } { = } { - } 3 7 . 4 7 ~ \mathrm { J /m o l{ \cdot }K }$  

At $273K$, the equilibrium constant of the above reaction is 1. Using Kirchhoff's law, the equilibrium constant at $268K$ can be calculated as follows:  

$K _ { 2 } = 1 . 0 4 3 = [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) + 2 n ( \mathrm { N a C l } ) ] / n ( \mathrm { H } _ { 2 } \mathrm { O } , l )$  

<Checkpoint>[$K _ { 2 } = 1 . 0 4 3 = [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) + 2 n ( \mathrm { N a C l } ) ] / n ( \mathrm { H } _ { 2 } \mathrm { O } , l )$ ](2)</Checkpoint>  

Additionally, since $\Delta H = 0$ throughout the process, we have:  

$0 = 5 0 7 1 . 2 [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) - n _ { 0 } ( \mathrm { H } _ { 2 } \mathrm { O } , l ) ] + C \times ( 2 6 8 \mathrm { K } - 273K)$  

<Checkpoint>[$0 = 5 0 7 1 . 2 [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) - n _ { 0 } ( \mathrm { H } _ { 2 } \mathrm { O } , l ) ] + C \times ( 2 6 8 \mathrm { K } - 273K)$](2)</Checkpoint>  

Solving the two equations simultaneously yields: $n ( \mathrm { H } _ { 2 } \mathrm { O } , s ) = 4 . 9 1 ~ \mathrm { m o l } ,$ , $n ( \mathrm { N a C l } ) = 0 . 1 3 3 \mathrm { m o l }$  

<Checkpoint>[$n ( \mathrm { H } _ { 2 } \mathrm { O } , s ) = 4 . 9 1 ~ \mathrm { m o l } ,$ , $n ( \mathrm { N a C l } ) = 0 . 1 3 3 \mathrm { m o l }$](2)</Checkpoint>  

The required mass of NaCl is $7.78 g$, and the mass of ice is $88.5 g$.
在本题中，由于NaCl质量较大，直接使用凝固点降低定律会造成误差！

$\mathrm { H } _ { 2 } \mathrm { O } ( l ) = \mathrm { H } _ { 2 } \mathrm { O } ( s )$ 

$\Delta _ { \mathrm { r } } C _ { \mathrm { m } } { = } { - } 3 7 . 4 7 ~ \mathrm { J /m o l{ \cdot }K }$

在$273K$ 下，上述反应的平衡常数为1。利用基尔霍夫定律可以下面计算268K下的平衡常数：

$K _ { 2 } = 1 . 0 4 3 = [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) + 2 n ( \mathrm { N a C l } ) ] / n ( \mathrm { H } _ { 2 } \mathrm { O } , l )$ 

<Checkpoint>[$K _ { 2 } = 1 . 0 4 3 = [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) + 2 n ( \mathrm { N a C l } ) ] / n ( \mathrm { H } _ { 2 } \mathrm { O } , l )$ ](2)</Checkpoint>

此外，整个过程中 $\Delta H = 0$ ，因此有：

$0 = 5 0 7 1 . 2 [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) - n _ { 0 } ( \mathrm { H } _ { 2 } \mathrm { O } , l ) ] + C \times ( 2 6 8 \mathrm { K } - 273K)$

<Checkpoint>[$0 = 5 0 7 1 . 2 [ n ( \mathrm { H } _ { 2 } \mathrm { O } , l ) - n _ { 0 } ( \mathrm { H } _ { 2 } \mathrm { O } , l ) ] + C \times ( 2 6 8 \mathrm { K } - 273K)$](2)</Checkpoint>

联立二式解得：$n ( \mathrm { H } _ { 2 } \mathrm { O } , s ) = 4 . 9 1 ~ \mathrm { m o l } ,$ ， $n ( \mathrm { N a C l } ) = 0 . 1 3 3 \mathrm { m o l }$

<Checkpoint>[$n ( \mathrm { H } _ { 2 } \mathrm { O } , s ) = 4 . 9 1 ~ \mathrm { m o l } ,$ ， $n ( \mathrm { N a C l } ) = 0 . 1 3 3 \mathrm { m o l }$](2)</Checkpoint>

所需NaCl的质量为$7.78 g$，冰的质量为$88.5 g$