Answer: The mass percent of water in the original air sample is 4.16 %
Explanation:
We are given:
Volume of moist air = 12.0 L
Volume of dry air = 11.50 L
Volume of water lost = (12.0 - 11.50) L = 0.50 L
Here, the percent by mass ratio will be equal to the percent by volume ratio as the number of moles is directly related to the volume.
To calculate the volume percentage of water in the air, we use the equation:
[tex]\text{Volume percent of water}=\frac{\text{Volume of water}}{\text{Volume of air}}\times 100[/tex]
Putting values in above equation, we get:
[tex]\text{Volume percent of water}=\frac{0.50L}{12.00L}\times 100\\\\\text{Volume percent of water}=4.16\%[/tex]
Hence, the mass percent of water in the original air sample is 4.16 %
