Question

What is the vapor pressure in mm

 Hg of a solution prepared by dissolving 25.0 g

 of ethyl alcohol (C

2

H

5

OH)

 in 100.0 g

 of water at 25

∘

C?

The vapor pressure of pure

water is 23.8mmHg,

 and the vapor pressure of ethyl alcohol is 61.2 mmHg

at 25

∘

C.

(b) What is the vapor pressure of the solution if 25.0 g

 of water is dissolved in 100.0 g

 of ethyl alcohol at 25

∘

C?

Answer

Answer 1

(a)

Molar mass of water $=18.015$

Molar mass of Ethyl alcohol$=46.07$

Moles of water $=\frac{100}{18.015}=5.551$

Moles of $C_2H_5OH=\frac{25}{46.07}=0.543$

Total moles $=6.094moles$

$P_{water}=\frac{5.551}{6.094}×23.8=21.679$

$P_{alcohol}=\frac{0.543}{6.094}×61.2=5.453$

Vapor pressure of solution $=21.679+5.453$

$=27.132mmHg$

(b)

Moles of water $=\frac{25}{18.015}=1.388$

Moles of ethyl alcohol$=\frac{100}{46.07}=2.171$

Total moles $=3.559$

$P_{water}=\frac{1.388}{3.559}×23.8=9.282$

$P_{Alcohol}=\frac{2.171}{3.559}×61.2=37.332$

Vapor pressure of solution

$=9.282+37.332$

$=46.614mmHg$