The vapor pressure of carbon disulfide is 355.6 torr at 25°C. What is the vapor pressure of a solution prepared by dissolving 10.60 g naphthalene (C10H8, Molar Mass = 128.2 g/mol) in 155 mL CS2 liquid (Molar Mass = 76.14 g/mol, density = 1.261 g/mL)? Assume the solution obeys Raoult's law, and treat naphthalene as a nonvolatile solute.

Answers

Answer 1

Answer:

344.5764 torr

Explanation:

Molar mass of naphthalene = 128.2g/mol

Mass of naphthalene = 10.60 g

Carbon disulfide:

Molar mass= 76.14g/mol ;

volume = 155mL ;

density = 1.261 g/mL

Vapour pressure = 355.6 torr

Number of moles = mass / molar mass

CS2:

Mass = density × volume

Number of moles = (density × volume) / molar mass

Number of moles = (1.261 * 155) / 76.14 = (195.455 / 76.14) = 2.567 moles of CS2

Number of moles of C8H10:

Number of moles = 10.60 / 128.2 = 0.083 C8H10

Total number of moles :

2.567 + 0.083 = 2.65 moles

Mole fraction of each compound in solution :

CS2 :

2.567 / 2.65 = 0.969

C8H10:

0.083 / 2.65 = 0.031

According to Raoult's:

Psolution = Xsolvent × Posolvent

CS2 = solvent

Xsolvent = Mole fraction of solvent

Posolvent = Vapour pressure of pure solvent

Psolution = 0.969 × 355.6 torr = 344.5764 torr

Answer 2

The vapor pressure of the solution would be as follows:

[tex]344.5764[/tex] torr

Given that,

Vapor pressure of Carbon Disulfide [tex]= 355.6[/tex] torr

[tex]volume = 155mL ;[/tex]

[tex]density = 1.261 g/mL[/tex]

Naphthalene's mass [tex]= 10.60 g[/tex]

Naphthalene's molar mass [tex]= 128.2g/mol[/tex]

Now,

We know that

Number of moles [tex]= mass/molar mass[/tex]

Mass [tex]= density[/tex] × [tex]volume[/tex]

[tex]Number of moles =[/tex] [tex](density[/tex] × [tex]volume) / molar mass[/tex]

So,

Number of moles of Carbon Disulfide [tex]= (1.261[/tex] × [tex]155) / 76.14[/tex]

[tex]= (195.455 / 76.14)[/tex]

[tex]= 2.567[/tex] moles of Carbon Disulfide

Number of moles of Naphthalene:

Number of moles [tex]= 10.60 / 128.2[/tex]

[tex]= 0.083[/tex]

Now,

Total number of moles :

[tex]2.567 + 0.083[/tex]

[tex]= 2.65[/tex] moles

Mole fraction of each compound in solution :

Carbon Disulfide:

2.567 / 2.65

[tex]= 0.969[/tex]

Naphthalene

0.083 / 2.65

[tex]= 0.031[/tex]

According to Raoult's:

P[tex]solution = Xsolvent[/tex] × [tex]Posolvent[/tex]

Carbon Sulfide = Solvent

[tex]Xsolvent =[/tex] Mole fraction of solvent

[tex]Posolvent =[/tex] Vapour  pressure of the pure solvent

[tex]Psolution[/tex] [tex]= 0.969[/tex] × [tex]355.6 torr[/tex]

[tex]= 344.5764[/tex] torr

Thus, "[tex]344.5764[/tex] torr" is the correct answer.

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Related Questions

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Answers

Answer:

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Answers

Answer:

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Answers

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Answers

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Answers

Answer:

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Explanation:

Here, we want to select which of the options are NMR inactive. For an isotope to be NMR inactive, then the net nuclear spin is zero.

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Answers

Answer:

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Answers

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Answers

Answer:

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Answers

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Answers

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Answers

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Answers

Answer:

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Answers

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Answers

Answer:

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Which of the following statements is true about the arrangement of particles in matter?
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A. The particles in a gas are closer together than the particles in a liquid
D. The particles in a solid are spread farther apart than the particles in a liquid
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Answers

Answer:

C. The particles in a gas are spread farther apart than the particles in a solid

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Answers

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Answers

Answer:

See explanation

Explanation:

(a)

From the data provided;

Mass of methanol =11.0 g

Mass of solvent (H2O) =100 g = 0.1 kg

Molar mass of CH3OH =32.04 g/mol

Number of moles of solute = 0.343 mol

molality(m) of methanol = 3.43 m

Also,

Mass of ethanol =22.0 g

Mass of solvent (H2O) =200 g = 0.2 kg

Molar mass of C2H5OH =46.068 g/mol

Number of moles of solute = 0.477 mol

molality(m) of ethanol= 2.38 m

The greater the concentration of a solution, the higher the freezing point of solvent will be depressed. Therefore, higher molality, leader to a greater decrease of the freezing point.

Therefore, the freezing point of methanol CH3OH/H2O is lower than the freezing point of ethanol.

b)

Mass of solvent = 1 kg

Mass of water = 20g or 0.02 Kg

Molar mass of water = 18 gmol-1

Number of moles of solute = 20g/18 gmol-1 = 1.11 moles

Molality= 1.11 moles/1 kg = 1.11 m

Also;

Mass of ethanol = 20 g = 0.02 kg

Mass of solvent = 1.00 Kg

Molar mass of solute = 46 gmol-1

Number of moles of solute = 20g/ 46 gmol-1 = 0.43 moles

Molality= 0.43 moles/ 1 kg = 0.43 m

The greater the concentration of a solution, the higher the freezing point of solvent will be depressed. Therefore, higher molality, leader to a greater decrease of the freezing point.

Therefore, the freezing point of water H2O/CH3OH is lower than the freezing point of ethanol.

Not divisible
O atom
O compound
O mixture
O solution

Answers

Answer:

mixture or atom

Explanation:

I HOPE IT WILL HELP YOU

Explanation:

and ,answer is mixture.

(01.04 LC)
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1) Gas
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4) Solid

Answers

the answer is liquid ! hope this helped :)
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