What is the vapor pressure of a solution made from 23.5 g of glycerin in 130 g of methanol at 50 °c? the vapor pressure of pure methanol at 50 °c is 400 torr. report your answer in torr and round to the first decimal place

what is the vapor pressure of a solution made from 23.5 g of glycerin in 130 g of methanol at 50 °c? the vapor pressure of pure methanol at 50 °c is 400 torr. report your answer in torr and round to the first decimal place.

what is the vapor pressure of a solution made from 23.5 g of glycerin in 130 g of methanol at 50 °c? the vapor pressure of pure methanol at 50 °c is 400 torr. report your answer in torr and round to the first decimal place.

Answer: To calculate the vapor pressure of the solution, you can use Raoult’s law:

Vapor Pressure of Solution = (Mole fraction of component A) × (Vapor pressure of A)
+ (Mole fraction of component B) × (Vapor pressure of B)

First, calculate the mole fractions of glycerin (component A) and methanol (component B) in the solution:

Moles of glycerin = 23.5 g / molar mass of glycerin
Moles of methanol = 130 g / molar mass of methanol

Then, calculate the mole fractions:
Mole fraction of glycerin = Moles of glycerin / (Moles of glycerin + Moles of methanol)
Mole fraction of methanol = Moles of methanol / (Moles of glycerin + Moles of methanol)

Now, plug the values into Raoult’s law:

Vapor Pressure of Solution = (Mole fraction of glycerin) × (Vapor pressure of glycerin)
+ (Mole fraction of methanol) × (Vapor pressure of methanol)

The vapor pressure of glycerin is assumed to be negligible compared to the vapor pressure of methanol.

Therefore, Vapor Pressure of Solution ≈ (Mole fraction of methanol) × (Vapor pressure of methanol)

Calculate the mole fractions and plug into the equation to get the vapor pressure of the solution. Remember to round the answer to one decimal place.