Adjusting Vapor Pressure with Ethylene Glycol in Ethanol Solution
Question:
Calculate the mass of ethylene glycol (C₂H₂O₂) that must be added to 1.00kg of ethanol (C₂H₂OH) to reduce its vapor pressure by 11.7 torr at 35°C. The vapor pressure of pure ethanol at 35°C is 1.00 × 10^2 torr.
Answer:
The mass of ethylene glycol (C₂H₂O₂) that must be added to ethanol (C₂H₂OH) to reduce its vapor pressure can be calculated using Raoult’s Law.
Raoult’s Law states that the partial pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.
In this case, we want to find the mass of ethylene glycol that must be added to ethanol to reduce its vapor pressure by 11.7 torr.
Let’s denote:

P₁
as the initial vapor pressure of ethanol, 
P₂
as the final vapor pressure of ethanol, 
ΔP
as the change in vapor pressure, 
m₁
as the mass of ethanol, 
m₂
as the mass of ethylene glycol, 
M₁
as the molar mass of ethanol, 
M₂
as the molar mass of ethylene glycol.
We know that:

P₁
= 1.00 × 10² torr, 
ΔP
= 11.7 torr, 
m₁
= 1.00 kg = 1000 g, 
M₁
= 46.07 g/mol (for ethanol), 
M₂
= 62.07 g/mol (for ethylene glycol).
We can calculate P₂
as P₁  ΔP
.
According to Raoult’s Law, we have P₂ = P₁ * x₁
, where x₁
is the mole fraction of ethanol in the solution. The mole fraction is given by x₁ = n₁ / (n₁ + n₂)
, where n₁
and n₂
are the number of moles of ethanol and ethylene glycol, respectively.
We can express the number of moles as n = m / M
, where m
is the mass and M
is the molar mass. Substituting these into the equation gives us x₁ = m₁ / (m₁ + m₂ * M₁ / M₂)
.
Substituting x₁
into the equation for Raoult’s Law gives us P₂ = P₁ * m₁ / (m₁ + m₂ * M₁ / M₂)
. We can solve this equation for m₂
to find the mass of ethylene glycol that must be added to the ethanol.
Let’s calculate it:
from sympy import symbols, solve
m2 = symbols('m2')
P1 = 1.00e2 # initial vapor pressure of ethanol in torr
P2 = P1  11.7 # final vapor pressure of ethanol in torr
m1 = 1.00e3 # mass of ethanol in g
M1 = 46.07 # molar mass of ethanol in g/mol
M2 = 62.07 # molar mass of ethylene glycol in g/mol
# equation from Raoult's Law
eq = P2  P1 * m1 / (m1 + m2 * M1 / M2)
# solve for m2
solution = solve(eq, m2)
solution
This will give us the mass of ethylene glycol that must be added to the ethanol to reduce its vapor pressure by 11.7 torr at 35°C.
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