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katovenus [111]
3 years ago
7

171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in o

ne liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?
Chemistry
1 answer:
Fantom [35]3 years ago
4 0

The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?

a. 16mm Hg

b. 24mm Hg  

c. 20mm Hg  

d. 12mm Hg

Answer: The vapor pressure (VP) of this solution at 25^{o}C is closest to the value 24 mm Hg.

Explanation:

Given: Mass of sucrose = 171 g

Mass of water = 1 L = 1000 g

Vapor pressure of water = 24 mm Hg

As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.

Moles = \frac{mass}{molar mass}\\= \frac{1000 g}{18.02 g/mol}\\= 55.49 mol

Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.

Moles = \frac{mass}{molar mass}\\= \frac{171 g}{342 g/mol}\\= 0.5 mol

Total moles = 55.49 + 0.5 mol = 55.99 mol

Mole fraction of water is as follows.

Mole fraction = \frac{moles of water}{total moles}\\= \frac{55.49}{55.99}\\= 0.99

Formula used to calculate vapor pressure of the solution is as follows.

P_{i} = P^{o}_{i} \times \chi_{i}

where,

P_{i} = vapor pressure of component i over the solution

P^{o}_{i} = vapor pressure of pure component i

\chi_{i} = mole fraction of i

Substitute the values into above formula to calculate vapor pressure of water as follows.

P_{i} = P^{o}_{i} \times \chi_{i}\\= 24 mm Hg \times 0.99\\= 23.76 \\or 24 mm Hg\\

Thus, we can conclude that the vapor pressure (VP) of this solution at 25^{o}C is closest to the value 24 mm Hg.

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Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


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Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


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i) chemical equation:


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H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

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