Question
Asked Feb 6, 2020
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Carbon disulfide, CS2, is a volatile, flammable liquid. It has a vapor pressure of 400.0 mmHg at 28.0  degrees celsius and a normal boiling point of 46.5 degrees celsius.

A) What is the heat of  vaporization of this substance in KJ/mol?

B)At which pressure would CS2 have a boilingpoint of 75 degrees celsius?

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Expert Answer

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Given data:

Vapor pressure (P1) = 760 mmHg (normal boiling point pressure)

Normal boiling temperature (T1) = 46.5 ℃ = 319.65 K.

Vapor pressure (P2) = 400 mmHg

Final temperature (T2) = 28 ℃ = 301.15 K

Temperature (T3) = 75 ℃ = 348.15 K.

Formula:

Clausius-Clapeyron Equation is given by,

Chemistry homework question answer, step 2, image 1

A. Calculation for heat of vaporization using Clausius-Clapeyron Equation:

Chemistry homework question answer, step 3, image 1

...

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