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Tom [10]
3 years ago
8

Calculate the vapor pressure at 35ºC of a solution made by dissolving 20.2 g of sucrose (C12H22O11)in 60.5 g of water. The vapor

pressure of pure water at 35ºC is 42.2 mmHg. What is the vapor-pressure depressionof the solution(in units of mmHg)? (Sucrose is nonvolatile)
Chemistry
1 answer:
sukhopar [10]3 years ago
5 0

Answer:

P' = 41.4 mmHg → Vapor pressure of solution

Explanation:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - Vapor pressure of solution (P')

Xm = Mole fraction for solute (Moles of solvent /Total moles)

Firstly we determine the mole fraction of solute.

Moles of solute → Mass . 1 mol / molar mass

20.2 g . 1 mol / 342 g = 0.0590 mol

Moles of solvent → Mass . 1mol / molar mass

60.5 g . 1 mol/ 18 g = 3.36 mol

Total moles = 3.36 mol + 0.0590 mol = 3.419 moles

Xm = 0.0590 mol / 3.419 moles → 0.0172

Let's replace the data in the formula

42.2 mmHg - P' = 42.2 mmHg . 0.0172

P' = - (42.2 mmHg . 0.0172 - 42.2 mmHg)

P' = 41.4 mmHg

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What is the partial pressure of argon, par, in the flask? express your answer to three significant figures and include the appro
monitta
Given which are missing in your question:
the flask is filled with 1.45 g of argon at 25 C° 
So according to this formula (Partial pressure):
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first, we need n, and we can get by substitution by:
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The series which is in order of increasing boiling point is CH3CH2CH3 CH3COCH3 CH2CH2CH3OH

However, the boiling point of an organic substance is the temperature at which the vapor pressure of the liquid organic substance equals the pressure surrounding the liquid and the liquid changes into a vapor.

<h3>What are organic compounds?</h3>

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