On a particular summer day, the temperature is 30.0°C and the relative humidity is 79.7% (which means that the partial pressure of water vapor in the ambient air is 79.7% of the equilibrium vapor pressure of water). A sample of this air is placed in a 1.00 L flask, which is then closed and cooled to 14.0°C. At 30.0°C, the vapor pressure of water is 31.8 mmHg, and at 14.0°C is 12.0 mmHg. What is the mass (in grams) of water that still exists as vapor in the flask? Mass = g How much liquid water (in grams) condenses out? Mass = g Submit Answer Try Another Version 5 item attempts remaining

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter11: Liquids And Solids
Section: Chapter Questions
Problem 11.96QE
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On a particular summer day, the temperature is 30.0°C and the relative humidity is 79.7% (which means that the partial pressure of water vapor in the ambient air is 79.7% of the
equilibrium vapor pressure of water). A sample of this air is placed in a 1.00 L flask, which is then closed and cooled to 14.0°C. At 30.0°C, the vapor pressure of water is 31.8 mmHg,
and at 14.0°C is 12.0 mmHg.
What is the mass (in grams) of water that still exists as vapor in the flask?
Mass =
How much liquid water (in grams) condenses out?
Mass =
Submit Answer
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Transcribed Image Text:On a particular summer day, the temperature is 30.0°C and the relative humidity is 79.7% (which means that the partial pressure of water vapor in the ambient air is 79.7% of the equilibrium vapor pressure of water). A sample of this air is placed in a 1.00 L flask, which is then closed and cooled to 14.0°C. At 30.0°C, the vapor pressure of water is 31.8 mmHg, and at 14.0°C is 12.0 mmHg. What is the mass (in grams) of water that still exists as vapor in the flask? Mass = How much liquid water (in grams) condenses out? Mass = Submit Answer Try Another Version 5 item attempts remaining
Expert Solution
Step 1

The expression of ideal gas equation is shown below:

PV = nRT

Where;

P = pressure

V = volume

n = moles

T = temperature

Step 2

Conversion of pressure of H2O mmHg to atm is shown below:

Pressure of H2O = 31.8 mmHg×79.7100=25.345mm Hg=0.03335 atm

Step 3

The calculation of moles at 300C is shown below:

n =PVRT=0.03335×10.0821×303.15=0.00134 mol

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