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.

The vapor pressure of the solution would be as follows:

torr

Given that,

Vapor pressure of Carbon Disulfide torr

Naphthalene's mass

Naphthalene's molar mass

Now,

We know that

Number of moles

Mass ×

×

So,

Number of moles of Carbon Disulfide ×

moles of Carbon Disulfide

Number of moles of Naphthalene:

Number of moles

Now,

Total number of moles :

moles

Mole fraction of each compound in solution :

Carbon Disulfide:

2.567 / 2.65

Naphthalene

0.083 / 2.65

According to Raoult's:

P ×

Carbon Sulfide = Solvent

Mole fraction of solvent

Vapour  pressure of the pure solvent

×

torr

Thus, " torr" is the correct answer.

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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

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