The decrease in vapor pressure of an aqueous solution that contains 45.0% ethylene glycol (C₂H₄(OH)₂) by mass is _____ mmHg. (Given; vapor pressure of pure water at 100.0˚C is 760.0 mmHg) A) 656 B) 569 C) 614 D) 164 E) 416

Question

Quizplus · Accepted Answer

To find the decrease in vapor pressure, we need to use Raoult's Law.

The vapor pressure of the solution, P(solution), is equal to the mole fraction of the solvent, water (x(water)), multiplied by the vapor pressure of pure water (P(water)).

P(solution) = x(water) * P(water)

Since we are given the mass percentage of ethylene glycol, we can find the mole fraction of water as follows:

mass fraction of water = 1 - 0.45 = 0.55
mole fraction of water = (0.55 g water / 18.0 g/mol) / [(0.55 g water / 18.0 g/mol) + (0.45 g ethylene glycol / 62.07 g/mol)]
= 0.879

Now, we can find the vapor pressure of the solution:

P(solution) = 0.879 * 760.0 mmHg
= 668.8 mmHg

To find the decrease in vapor pressure, we subtract the vapor pressure of the solution from the vapor pressure of pure water:

ΔP = P(water) - P(solution)
= 760.0 mmHg - 668.8 mmHg
= 91.2 mmHg

Therefore, the decrease in vapor pressure of the aqueous solution is 91.2 mmHg, which corresponds to choice (C).

The Decrease in Vapor Pressure of an Aqueous Solution That