Glycerol is a nonvolatile water soluble material. It's density is 1.25g/ml.
Predict the vapor pressure of a solution of 2.40E2 ml of glycerol, in 385ml of water, at the normal boiling pt of water.
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To predict the vapor pressure of this solution, we need to apply Raoult's law. According to Raoult's law, the vapor pressure of a component in an ideal solution is directly proportional to its mole fraction in the solution.
First, let's calculate the number of moles of glycerol and water in the solution:
- Glycerol:
The volume of glycerol is given as 2.40E2 ml. To convert it to liters, we divide by 1000:
Volume of glycerol = 2.40E2 ml ÷ 1000 = 0.24 L
Now, using the density of glycerol (1.25 g/mL), we can calculate the mass of glycerol:
Mass of glycerol = Volume of glycerol × Density of glycerol
Mass of glycerol = 0.24 L × 1.25 g/mL = 0.30 g
Next, we can calculate the number of moles of glycerol using its molar mass, which is 92.1 g/mol:
Number of moles of glycerol = Mass of glycerol ÷ Molar mass of glycerol
Number of moles of glycerol = 0.30 g ÷ 92.1 g/mol ≈ 0.00326 mol
- Water:
The volume of water is given as 385 ml. Converting it to liters:
Volume of water = 385 ml ÷ 1000 = 0.385 L
Now, we can use the mole fraction to determine the vapor pressure of the solution. The mole fraction is calculated by dividing the number of moles of glycerol by the total number of moles (glycerol + water).
Total number of moles = Number of moles of glycerol + Number of moles of water
Total number of moles = 0.00326 mol + 0.385 L
Finally, applying Raoult's law:
Vapor pressure of the solution = Mole fraction of glycerol × Vapor pressure of pure water
The vapor pressure of pure water at its normal boiling point is standardly defined as 1 atm.
Now you can substitute the values into the equation to get your final answer.