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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Answers

Answer:

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Explanation:

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Answers

Answer:

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Answers

Answer:

46.6 laps

Explanation:

Step 1: Given data

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1056 ft × (1 km / 3280.84 ft) = 0.3219 km

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Explanation:

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Answers

Answer:

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Explanation:

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Answers

Answer:

e. None of the answer choices

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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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Answers

Answer:

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Answers

Answer:

4.54x10^-19 J

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Answers

Answer:

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Answer:

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Answer:

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Answer:

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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

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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

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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

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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.

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Answers

Answer:

69.7% is percent yield

Explanation:

Based on the reaction:

3Ca(NO3)2(aq) + 2Na3PO4(aq) → Ca3(PO4)2(s) + 6NaNO3(aq)

2 moles of Na3PO4 react producing 6 moles of NaNO3.

As 24.2 moles of Na3PO4 react, theoretical moles of NaNO3 produced are:

24.2 moles Na3PO4 * (6 moles NaNO3 / 2 moles Na3PO4) =

72.6 moles of NaNO3

As there are produced 50.6 moles of NaNO3, percent yield is:

50.6 moles NaNO3 / 72.6 moles NaNO3 =

69.7% is percent yield

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Answers

Answer:

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If your front lawn is 25.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1350 new snowflakes every minute, how much snow, in kilograms, accumulates on your lawn per hour? Assume an average snowflake has a mass of 1.90 mg.

Answers

Answer:

19.17 kg

Explanation:

calculate area of lawn

25  * 20  =  500  sq ft

calculate how many flakes fall per minute

500  *  1150  =   575 , 000  flakes/minute

convert per minute rate to per hour rate

575 , 000  * 60  = 34 ,500 , 000  flakes/hour

now you know how many flakes feel in that one hour so we can now weigh them

34 , 500 , 000  / 1.8  = 19 , 166 , 667  mg (rounded)

convert to kilograms

19 , 166 , 667 /  100 , 000  = 19.17  kg (rounded)

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