11.4k views
0 votes
Determine the vapor pressure in torr of the solution when 43.78 g of a non-volatile, non-electrolyte substance X with molar mass of 210 g is dissolved at in 100 g of ethanol at 298K. Please note that ethanol has a molar mass of 46.07 g/mol and vapor pressure of 58.9 torr at 298K.

User Sibvic
by
4.3k points

2 Answers

1 vote

Final answer:

The vapor pressure of the solution, when a non-volatile substance is dissolved in ethanol, is determined using Raoult's law and is found to be 53.7 torr at 298K.

Step-by-step explanation:

To determine the vapor pressure of the solution, we first need to find the mole fraction of ethanol in the solution using Raoult's law. We can calculate the moles of the substance X and ethanol, then derive the mole fraction of ethanol, and finally compute the new vapor pressure of the solution.

First, calculate the moles of substance X using its mass (43.78 g) and molar mass (210 g/mol).

Moles of substance X = 43.78 g / 210 g/mol = 0.2085 mol

Now, calculate the moles of ethanol using its mass (100 g) and molar mass (46.07 g/mol).

Moles of ethanol = 100 g / 46.07 g/mol = 2.171 mol

The total moles in the solution would be the sum of moles of substance X and ethanol, but since substance X is non-volatile and does not contribute to the vapor pressure, only the moles of ethanol are considered for the vapor pressure calculation.

Now, calculate the mole fraction of ethanol (Xethanol) in the solution:

Mole fraction of ethanol (Xethanol) = Moles of ethanol / (Moles of ethanol + Moles of substance X)

Xethanol = 2.171 mol / (2.171 mol + 0.2085 mol)

Xethanol = 2.171 mol / 2.3795 mol

Xethanol = 0.9125

To calculate the vapor pressure of the solution (Psolution), multiply the mole fraction of ethanol (Xethanol) by the vapor pressure of pure ethanol (Pethanol).

Vapor pressure of the solution (Psolution) = Xethanol × Pethanol

Psolution = 0.9125 × 58.9 torr

Psolution = 53.7 torr

User Andi Giga
by
5.4k points
2 votes

Answer:

The vapor pressure in torr of the solution is 53.7

Step-by-step explanation:

Vapor pressure lowering is the colligative property that is solved with this formula:

ΔP = P° . Xm

ΔP = Vapor pressure of pure solvent (P°) - vapor pressure of solution

Xm = mole fraction for solute.

Let's solve the mole fraction (moles of solute / total moles)

Total moles = moles of solute + moles of solvent

Moles of solute = 43.78 g / 210 g/mol = 0.208 moles

Moles of solvent = 100 g / 46.07 g/mol = 2.17 moles

Total moles = 2.17 mol + 0.208 mol → 2.378 moles

Mole fraction for solute = 0.208 / 2.378 = 0.087

Let's replace the data in the main formula.

P° = 58.9 Torr

Xm = 0.087

ΔP = P° . Xm

58.9 Torr - Vapor pressur of solution = 58.9 Torr . 0.087

Vapor pressure of solution = - (58.9 Torr . 0.087 - 58.9 Torr)

Vapor pressure of solution = 53.7 Torr

User FunctorPrototype
by
4.9k points