Known:
1. The triple point of water is $273.16 \mathrm{~K}$, and the freezing point is $273.15 \mathrm{~K}$.
2. The relationship between the freezing point and pressure is $\Delta \mathrm{T} / \Delta \mathrm{P}=-7.432 \times 10^{-8} \mathrm{~K} / \mathrm{Pa}$.
3. At the triple point, the vapor pressure of water is $610.48 \mathrm{~Pa}$.
4. The volume fractions of nitrogen and oxygen in the air are $79 \%$ and $21 \%$, and their Henry's constants are $1.035 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1} / \mathrm{kPa}$ and $2.154 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1}/ \mathrm{kPa}$, respectively.
5. The atmospheric pressure is $101325 \mathrm{~Pa}$, and the acceleration due to gravity is taken as $9.8 ~m/s^2$.

Based on the known information, the problems to be calculated are:

(a). The numerical value of the freezing point depression constant of water when all units are in SI units (keep two significant figures).

(b). Place a cup of water saturated with air on a mountain top at an altitude of $5000 ~m$, what is the freezing point of this cup of water (in K, keep five significant figures)?

The following options all contain two values, the first is the answer to question (a), and the second is the answer to question (b). The option with both answers correct is (calculation error less than 0.1% is considered correct):
已知：
1. 水的三相点为$273.16 \mathrm{~K}$，凝固点为$273.15 \mathrm{~K}$。
2. 凝固点与压力的关系为$\Delta \mathrm{T} / \Delta \mathrm{P}=-7.432 \times 10^{-8} \mathrm{~K} / \mathrm{Pa}$。
3. 三相点时, 水的蒸气压为 $610.48 \mathrm{~Pa}$。
4. 空气中氮气与氧气的体
积分数为 $79 \%$ 与 $21 \%$, 其亨利常数分别为 $1.035 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1} / \mathrm{kPa}$ 与 $2.154 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1}/ \mathrm{kPa}$。
5. 大气压为 $101325 \mathrm{~Pa}$，重力加速度取$9.8 ~m/s^2$。

根据已知信息，需要求算的问题为：

(a). 单位均取国际单位时，水的凝固点下降常数的数值（保留两位有效数字）

(b). 将一杯饱和了空气的水置于海拔$5000 ~m$的山顶，这杯水的凝固点（单位为K，小数点后保留两位有效数字）。

下列选项均含有两个数值，第一个为问题(a)的答案，第二个为问题(b)的答案，这两个答案均正确的选项是（计算误差小于0.1%视为正确）：
All other options are incorrect
1.8541，273.13
1.8541，273.15
1.8541，273.14
1.9542，273.13
1.9542，273.14
1.9542，273.15
2.0542，273.15
2.0542，273.14
2.0542，273.13
2.1542，273.13
2.1542，273.14
2.1542，273.15
其他选项均不正确
1.8541，273.13
1.8541，273.15
1.8541，273.14
1.9542，273.13
1.9542，273.14
1.9542，273.15
2.0542，273.15
2.0542，273.14
2.0542，273.13
2.1542，273.13
2.1542，273.14
2.1542，273.15
Problem (a):

The freezing point depression consists of two parts: the relationship between pressure and freezing point, and the colligative property of dilute solutions of air dissolved in water leading to freezing point depression.

<Checkpoint>[The freezing point depression consists of two parts: the relationship between pressure and freezing point, and the colligative property of dilute solutions of air dissolved in water leading to freezing point depression](2)</Checkpoint>

Freezing point change caused by external pressure:

$\Delta T_p = -7.423 \times 10^{-8} \times (101325 - 610.48) = -7.485 \times 10^{-3} ~K$

<Checkpoint>[Freezing point change caused by external pressure $\Delta T_p = -7.485 \times 10^{-3} ~K$](1)</Checkpoint>

According to Henry's law, the molar concentration $c$ of dissolved gas is proportional to the partial pressure $p$ of the gas

$c = k_H \cdot p$

<Checkpoint>[Henry's law $c = k_H \cdot p$](1)</Checkpoint>

Therefore, the concentration of air in saturated water is:

$c = c_{N_2} + c_{O_2} = (1.035 \times 10^{-5}\times 79\% + 2.154\times 10^{-5}\times 21\%)\times 101.325 = 1.287 \times 10^{-3} ~mol/L$

<Checkpoint>[Concentration of air in saturated water $c = c_{N_2} + c_{O_2} = 1.287 \times 10^{-3} ~mol/L$](1)</Checkpoint>

For dilute aqueous solutions, the molality $b$ is numerically approximately equal to the molarity $c$

<Checkpoint>[Dilute aqueous solutions, the molality $b$ is numerically approximately equal to the molarity $c$](1)</Checkpoint>

Freezing point change caused by colligative properties

$\Delta T_c = K_f \cdot b$

<Checkpoint>[Colligative properties $\Delta T_c = K_f \cdot b$](1)</Checkpoint>

$K_f = \Delta T_c/b = (\Delta T - \Delta T_p)/c = (0.01 - 7.485 \times 10^{-3})K/(1.287 \times 10^{-3})mol/L = 1.9542 K/kg\cdot mol$

<Checkpoint>[$K_f= 1.9542 K/kg\cdot mol$](2)</Checkpoint>

Problem (b):

First, it is necessary to calculate the change in pressure with altitude according to the Boltzmann distribution:

$\ln(p_1/p_2) = -Mg\Delta h/RT$

<Checkpoint>[Boltzmann distribution $\ln(p_1/p_2) = -Mg\Delta h/RT$](1)</Checkpoint>

Where $M$ represents the relative molecular mass of the gas.

The partial pressures of gases at $5000 ~m$ are calculated as:

$p_{N_2,5000} = 46 ~kPa, p_{O_2,5000} = 11 ~kPa$
(Since the gravitational acceleration is taken as 9.8, only two significant figures are retained subsequently)

<Checkpoint>[Partial pressures of gases at $5000 ~m$ $p_{N_2,5000} = 46 ~kPa, p_{O_2,5000} = 11 ~kPa$](1)</Checkpoint>

At this time, the freezing point change caused by colligative properties is

$\Delta T_c = K_f \cdot b = 2.0\times(1.035 \times 10^{-5} \times 46 + 2.154 \times 10^{-5} \times 11 = 1.4\times 10^{-3} ~K)$

<Checkpoint>[Freezing point change caused by colligative properties $\Delta T_c = 1.4\times 10^{-3} ~K)$](1)</Checkpoint>

The freezing point change caused by pressure is

$\Delta T_p = -7.423 \times 10^{-8} \times (46 + 11)\times 1000 = -4.2 \times 10^{-3} ~K$

<Checkpoint>[Freezing point change caused by pressure $\Delta T_p = -4.2 \times 10^{-3} ~K$](1)</Checkpoint>

Therefore, the total freezing point change is

$\Delta T = -\Delta T_p + \Delta T_c = 5.6 \times 10^{-3} ~K$

The new freezing point is

$T = 273.16 - 5.6 \times 10^{-3} = 273.15 ~K$

<Checkpoint>[New freezing point $T = 273.15 ~K$](1)</Checkpoint>

Therefore, answer G is correct.
问题(a):

凝固点的降低由两部分组成：压力与凝固点的关系，以及空气溶于水的稀溶液依数性导致的凝固点降低。

<Checkpoint>[凝固点的降低由两部分组成：压力与凝固点的关系，以及空气溶于水的稀溶液依数性导致的凝固点降低](2)</Checkpoint>

由外压引起的凝固点变化：

$\Delta T_p = -7.423 \times 10^{-8} \times (101325 - 610.48) = -7.485 \times 10^{-3} ~K$

<Checkpoint>[外压引起的凝固点变化$\Delta T_p = -7.485 \times 10^{-3} ~K$](1)</Checkpoint>

根据亨利定律，溶解气体的摩尔浓度$c$与气体分压$p$成正比

$c = k_H \cdot p$

<Checkpoint>[亨利定律$c = k_H \cdot p$](1)</Checkpoint>

因此求出饱和水中空气浓度：

$c = c_{N_2} + c_{O_2} = (1.035 \times 10^{-5}\times 79\% + 2.154\times 10^{-5}\times 21\%)\times 101.325 = 1.287 \times 10^{-3} ~mol/L$

<Checkpoint>[饱和水中空气浓度$c = c_{N_2} + c_{O_2} = 1.287 \times 10^{-3} ~mol/L$](1)</Checkpoint>

对于稀水溶液，质量摩尔浓度$b$在数值上约等于体积摩尔浓度$c$

<Checkpoint>[稀水溶液，质量摩尔浓度$b$在数值上约等于体积摩尔浓度$c$](1)</Checkpoint>

由依数性引起的凝固点变化

$\Delta T_c = K_f \cdot b$

<Checkpoint>[依数性$\Delta T_c = K_f \cdot b$](1)</Checkpoint>

$K_f = \Delta T_c/b = (\Delta T - \Delta T_p)/c = (0.01 -7.485 \times 10^{-3})K/(1.287 \times 10^{-3})mol/L =1.9542 K/kg\cdot mol$

<Checkpoint>[$K_f= 1.9542 K/kg\cdot mol$](2)</Checkpoint>

问题(b):

首先需要根据玻尔兹曼分布求出压力随海拔的变化：

$\ln(p_1/p_2) = -Mg\Delta h/RT$

<Checkpoint>[玻尔兹曼分布 $\ln(p_1/p_2) = -Mg\Delta h/RT$](1)</Checkpoint>

其中$M$表示气体的相对分子质量。

计算得$5000 ~m$处气体分压：

$p_{N_2,5000} = 46 ~kPa, p_{O_2,5000} = 11 ~kPa$
（由于重力加速度取9.8，后续只保留两位有效数字）

<Checkpoint>[$5000 ~m$处气体分压$p_{N_2,5000} = 46 ~kPa, p_{O_2,5000} = 11 ~kPa$](1)</Checkpoint>

此时由依数性引起的凝固点变化

$\Delta T_c = K_f \cdot b = 2.0\times(1.035 \times 10^{-5} \times 46 + 2.154 \times 10^{-5} \times 11 = 1.4\times 10^{-3} ~K)$

<Checkpoint>[依数性引起的凝固点变化$\Delta T_c = 1.4\times 10^{-3} ~K)$](1)</Checkpoint>

由压力引起的凝固点变化

$\Delta T_p = -7.423 \times 10^{-8} \times (46 + 11)\times 1000 = -4.2 \times 10^{-3} ~K$

<Checkpoint>[压力引起的凝固点变化$\Delta T_p = -4.2 \times 10^{-3} ~K$](1)</Checkpoint>

因此总凝固点变化

$\Delta T = -\Delta T_p + \Delta T_c = 5.6 \times 10^{-3} ~K$

新的凝固点

$T = 273.16 - 5.6 \times 10^{-3} = 273.15 ~K$

<Checkpoint>[新的凝固点 $T = 273.15 ~K$](1)</Checkpoint>

因此答案G正确。