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