Sucrose is table sugar and has a formula C12H22O11. The molar mao sucrose is 342.3 g/mol. What would be the vapor pressure of an aqueous sugar solution that has 250.0 grams of water and 30.0 grams of sugar dissolved in the water at 35.0??

 

a. 42.1mmhg

b. 23.6 mmhg

c 20.5mmhg

d 35.0mmhg

e 41.9 mmhg

 

To find out the vapor pressure in the given solution, follow the given steps

Step I: Find out the vapor pressure of pure water at 35.0? using the given formula as:

Vapor pressure of water any temperature = e^(20.386 - (5132 / (temperature + 273))

From the given question you have:

Temperature = 35.0?

Using this data in above equation you have:

Vapor pressure of water = e^(20.386 - (5132 / (35 + 273))

Vapor pressure of water = e^(20.386 - (5132/308)

Vapor pressure of water = e^(20.386 - 16.662)

Vapor pressure of water = e^3.724

Vapor pressure of water = 42.2 mm of Hg

Step II: Find out mole fraction of each component in the solution

Given:

Mass of sugar = 30.0 g

Molar mass of sugar = 342.3 g/mol

Number of moles of sugar = mass/molar mass = 30.0/342.3 = 0.0876 moles

Mass of water = 250.0 g

Molar mass of water = 18.0 g/mol

Number of moles of water = mass/molar mass = 250.0/18.0 = 13.89 moles

Mole fraction of water in solution = number of moles of water/total number of moles

Mole fraction of water in solution = 13.89/[13.89 + 0.0876] = 13.89/13.9776 = 0.9937

Step III: Using Raoult Law find out the vapor pressure of water in solution as:

Vapor pressure of water in solution = Mole fraction of water x vapor pressure of pure water at 35.0?

Vapor pressure of water in solution = 0.9937 x 42.2 mm of Hg = 41.9 mm of Hg

Vapor pressure of water in solution = 41.9 mm of Hg