Freezing point depression is a colligative property.
The formula that rules it is:
ΔT f = Kf * m
Where m is the molality of the solution => m = moles of solute / Kg of solvent
=> m = 0.31 mol / 0.175 kg = 1.771 m
And kf = 1.86 °C/m
=> ΔTf = 1.771 m * 1.86 °C / m = 3.3 °C
=> Freezing point of the solution = normal freezing point of water - 3.3 °C = 0 - 3.3°C = - 3.3°C.
Answer: -3.3 °C
<h2>Answer:</h2>
Option 3 is correct.
3. The molecular collisions are perfectly elastic.
<h2>Explanation:</h2>
According to the kinetic molecular theory, the molecular collisions are perfectly elastic, so the the molecules of a gas striking with the wall of container are elastic and they are exerting considerable pressure on the walls of container. This pressure is actually exerted against the walls is less for real gases as compared to ideal gases.
Answer:
23 kgs = 809.6 ounces
Explanation:
given:
1 kg = 2.2 ibs
therefore, 23 kgs = 2.2*23 Ibs = 50.6 Ibs
since, 1 Ib = 16 ounces
therefore, 23 kgs = 50.6*16 ounces = 809.6 ounces
<u>Answer:</u> The boiling point of water in Tibet is 69.9°C
<u>Explanation:</u>
To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:
![\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%3D%5Cfrac%7B%5CDelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)
= final pressure = 240. mmHg
= Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature or normal boiling point of water = ![100^oC=[100+273]K=373K](https://tex.z-dn.net/?f=100%5EoC%3D%5B100%2B273%5DK%3D373K)
= final temperature = ?
Putting values in above equation, we get:
![\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7B240%7D%7B760%7D%29%3D%5Cfrac%7B40700J%2Fmol%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B373%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D%5C%5C%5C%5C-1.153%3D4895.36%5B%5Cfrac%7BT_2-373%7D%7B373T_2%7D%5D%5C%5C%5C%5CT_2%3D342.9K)
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
Hence, the boiling point of water in Tibet is 69.9°C
