Question

# A solution is made by mixing 36.0 mL of ethanol, C2H6O, and 64.0 mL of water....

A solution is made by mixing 36.0 mL of ethanol, C2H6O, and 64.0 mL of water. Assuming ideal behavior, what is the vapor pressure of the solution at 20

```Vap pressure of ethanol at 20 degc is 59.3 mm Hg
Vap pressure of water at 20 degc is 17.5 mm Hg

Mol wt of ethanol is 46.07 g/mol
Mol wt of water is 18

Density of etanol is 0.789 gm / cc
Densiity of water 1 gm / cc

Given soln contains

Moles of ethanol is 36*0.789/46.07 which is 0.62
Moles of water is 64/18 which is 3.55 moles

Total moles in given soln 4.17

Mole fraction of ethanol = 0.62/4.17 which is 0.1486
Mole fraction of water = 3.55/4.17 which is 0.8514

By Raoult's law

Partial pressure of ethanol = 59.3*0.1486 which is 8.81 mm Hg
Partial pressure of water = 17.5*0.8514 which is 14.9 mm Hg

Therefore vap pressure of soln which is the sum of partial pressure of ethanol + and partial pressure of water
stands at 8.81+14.9  = 23.71 mm Hg
```

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