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Calculate the mass of propylene glycol (C3H8O2)

that must be added to 0.320 kg
of water to reduce the vapor pressure by 2.88 torr
at 40 ∘C
( PH2O
at 40 ∘C=55.3torr
).

1 Answer

5 votes

The mass of propylene glycol that must be added to 0.320 kg of water to reduce the vapor pressure by 2.88 torr at 40 °C is approximately 1284 g.

To calculate the mass of propylene glycol (C₃H₈O₂) needed to reduce the vapor pressure by 2.88 torr at 40 °C, we'll use Raoult's law and the given information.

Given Data:

Mass of water (m_water): 0.320 kg

Vapor pressure of pure water at 40 °C (P_H2O): 55.3 torr

Change in vapor pressure (ΔP): 2.88 torr

Steps:

Calculate Mole Fraction of Water (X_H2O):

X_H2O = ΔP / P_H2O

X_H2O = 2.88 torr / 55.3 torr

X_H2O ≈ 0.0521

Calculate Moles of Water (n_H2O):

n_H2O = m_water / Molar Mass_H2O

n_H2O = 0.320 kg / 18.01528 g/mol

n_H2O ≈ 17.77 mol

Calculate Moles of Propylene Glycol (n_C3H8O2):

Since X_H2O is the mole fraction of water, X_C3H8O2 (mole fraction of propylene glycol) is 1 - X_H2O.

X_C3H8O2 = 1 - X_H2O

X_C3H8O2 ≈ 0.9479

n_C3H8O2 = X_C3H8O2 * n_H2O

n_C3H8O2 ≈ 0.9479 * 17.77 mol

n_C3H8O2 ≈ 16.85 mol

Calculate Mass of Propylene Glycol (m_C3H8O2):

m_C3H8O2 = n_C3H8O2 * Molar Mass_C3H8O2

m_C3H8O2 ≈ 16.85 mol * 76.095 g/mol

m_C3H8O2 ≈ 1284 g

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