The normal boiling point for acetone is 56.5°c. at an elevation of 5300 ft the atmospheric pressure is 630. torr. what would be the boiling point of acetone (δhvap = 32.0 kj/mol) at this elevation? webassign will check your answer for the correct number of significant figures. °c what would be the vapor pressure of acetone at 24.0°c at this elevation?

The vapor pressure of acetona at 24.0 °C is 212.22 torr

Explanation:

Step 1: Data given

The normal boiling point for acetone is 56.5°C.

At an elevation of 5300 ft the atmospheric pressure is 630 torr

ΔHvap = 32.0 kJ/mol

Temperature = 24.0 °C

Step 2: The clausius clapeyron equation:

ln(p1/p2) = (∆H/R) (1/T2 - 1/T1)

Normal atmospheric pressure = 1 atm = 760 torr, so

ln(760/630) = ((32.000 J/mol) / (8.314 J/K*mol)) (1/T2 - 1/329.65 K).

T2 = 324.4 K = 51.25 °C.  This is the boiling point of acetone at 5300 ft.

The vapor pressure = atmospheric pressure at the boiling point for any liquid.

ln(Pv/630atm) = ((32.000 J/mol) / (8.314 J/K*mol)) (1/324.4 K - 1/297.15)

ln (Pv/630) = -13615388.3

Pv = 212.22 torr

The vapor pressure of acetona at 24.0 °C is 212.22 torr

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