Mass of liquid present in the container:
Yes, the liquid will be present, and of 1.2481 g
Calculation:
Given:
Volume = 1.5L
Temperature = 25 + 273 = 298K
P = 23.76 torr = 0.0312 atm
Using the ideal gas equation:
PV = nRT
Where:
making n subject of formula:
n = [tex]\frac{PV}{RT}[/tex]
By putting all the values in the given equation, we get,
n = 0.0312 x 1.5 / 0.0821 x 298
n = 0.0019 moles
So, 0.00189 are moles of water in the vapor
Now, subtract 1.25g of water from moles of water vapor, and we will get the mass of the liquid left,
1.25 - 0.0019 = 1.2481 g
Learn more about the ideal gas equation here,
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