
Interpretation:
The mole fraction of urea present in the two given solutions at equilibrium should be calculated.
Concept Introduction:
Molarity (M): The concentration for solutions is expressed in terms of molarity. Molarity is number of moles of the solute present in liter of the solution.
Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in
Mole fraction: Concentration of the solution can also expressed by mole fraction. Mole fraction is equal to moles of the component divided by total moles of the mixture.

Answer to Problem 12.132QP
The mole fraction of urea present in
The mole fraction of urea present in
The mole fraction of urea present in both beakers at equilibrium is
Explanation of Solution
Given: Beaker 1
Beaker 2
In order to calculate the mole fraction of urea first the mole of urea and the water moles in two given beaker should be determined.
Calculate moles of urea:
Calculate moles of water:
Calculate mole fraction of urea in each beaker:
Mole fraction of urea in beaker 1is as follows,
Mole fraction of urea in beaker 2is as follows,
At equilibrium the mole fractions of water in both beakers will be equal. According to Raoult’s Law the vapor pressure of water in each beaker will also be equal.
The number of moles transferred between the beakers in order to attain equilibrium is y.
Calculate mole fractions of urea at equilibrium:
In beaker 1
In beaker 2
The mole fraction of urea present in the two given solutions at equilibrium was calculated.
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Chapter 12 Solutions
CHEMISTRY (LL) W/CNCT >BI<
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