1. Increased motion in the particles in a liquid allows the particles to
2 answers:
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The complete question is as follows: 171 g of sucrose ( MW of 342, melting point 186 oC, boiling point very high, and vapor pressure is negligible) is dissolved in one liter of water at 25 oC. At 25 oC the vapor pressure of water is 24 mmHg. Which value is closest to the vapor pressure (VP) of this solution at 25 oC?
a. 16mm Hg
b. 24mm Hg
c. 20mm Hg
d. 12mm Hg
Answer: The vapor pressure (VP) of this solution at is closest to the value 24 mm Hg.
Explanation:
Given: Mass of sucrose = 171 g
Mass of water = 1 L = 1000 g
Vapor pressure of water = 24 mm Hg
As moles is the mass of substance divided by its molar mass. Hence, moles of water (molar mass = 18.02 g) is calculated as follows.
Similarly, moles of sucrose (molar mass = 342 g/mol) is as follows.
Total moles = 55.49 + 0.5 mol = 55.99 mol
Mole fraction of water is as follows.
Formula used to calculate vapor pressure of the solution is as follows.
where,
= vapor pressure of component i over the solution
= vapor pressure of pure component i
= mole fraction of i
Substitute the values into above formula to calculate vapor pressure of water as follows.
Thus, we can conclude that the vapor pressure (VP) of this solution at is closest to the value 24 mm Hg.
Answer:
Explanation:
Regardless of the type of gas, 1 mole at standard temperature and pressure (STP) occupies a volume of 22.4 liters. In this case the gas is helium (He).
We can set up a ratio.
Multiply by the given number of moles.
The moles of helium will cancel.
Multiply.
5.25 moles of helium gas at STP is 117.6 liters of helium.