Chapter 12, Problem 12.132QP

Interpretation Introduction

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.

$\text{Molarity = }\frac{\text{No}\text{. of moles of solute}}{\text{Volume of solution in L}}$

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in $\text{12}\text{g}$ of ${}^{\text{12}}\text{C}$.

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.

$\text{Mole fraction =}\frac{\text{moles of a component}}{\text{total moles}}$

Answer to Problem 12.132QP

The mole fraction of urea present in $50mLof0.10M$ solution is $\text{1}{\text{.8×10}}^{\text{-3}}$.

The mole fraction of urea present in $50mLof0.20M$ solution is $\text{3}{\text{.6×10}}^{\text{-3}}$.

The mole fraction of urea present in both beakers at equilibrium is $\text{2}\text{.7}\times {\text{10}}^{\text{-3}}$.

Explanation of Solution

Given: Beaker 1$=50mLof0.10M$

            Beaker 2$50mLof0.20M$

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:

$\begin{array}{c}\text{Molarity = }\frac{\text{Moles of solute}}{\text{Volume of solution in L}}\\ \text{Moles of urea in beaker 1}=\text{Molarity}\times \text{Volume of solution in L}\\ \text{=}\text{0}\text{.10}\text{mol/}\cancel{\text{L}}\text{×0}\text{.050}\cancel{\text{L}}\\ \text{=}\text{0}\text{.005 mol}\end{array}$

$\begin{array}{c}\text{Moles of urea in beaker 2}=\text{Molarity}\times \text{Volume of solution in L}\\ \text{=}\text{0}\text{.20}\text{mol/}\cancel{\text{L}}\text{×0}\text{.050}\cancel{\text{L}}\\ \text{=}\text{0}\text{.010 mol}\end{array}$

Calculate moles of water:

$\begin{array}{c}\text{Moles of water = 50mL×}\frac{\text{1g}}{\text{1mL}}\text{×}\frac{\text{1mol}}{\text{18}\text{.02g}}\\ \text{=}\text{2}\text{.8}\text{mol}\end{array}$

Calculate mole fraction of urea in each beaker:

Mole fraction of urea in beaker 1is as follows,

$\begin{array}{c}\text{Mole Fraction = }\frac{\text{0}\text{.0050}\text{mol}}{\text{0}\text{.0050}\text{mol+2}\text{.8}\text{mol}}\\ \text{=}\text{1}\text{.8}\times {\text{10}}^{\text{-3}}\end{array}$

Mole fraction of urea in beaker 2is as follows,

$\begin{array}{c}\text{Mole Fraction = }\frac{\text{0}\text{.010}\text{mol}}{\text{0}\text{.010}\text{mol+2}\text{.8}\text{mol}}\\ \text{=}\text{3}\text{.6}\times {\text{10}}^{\text{-3}}\end{array}$

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.

$\begin{array}{c}\text{At equilibrium}\\ {\text{X}}_{\text{1}}{\text{= X}}_2\\ \frac{\text{0}\text{.0050}\text{mol}}{\text{0}\text{.0050}\text{mol+2}\text{.8}\text{mol-y}}\text{ = }\frac{\text{0}\text{.010}\text{mol}}{\text{0}\text{.010}\text{mol+2}\text{.8}\text{mol+y}}\\ \left(\text{0}\text{.0050}\text{mol}\right)\left(\text{0}\text{.010}\text{mol+2}\text{.8}\text{mol+y}\right)=\left(\text{0}\text{.010}\text{mol}\right)\left(\text{0}\text{.0050}\text{mol+2}\text{.8}\text{mol-y}\right)\\ 0.014+0.0050y=0.028-0.010y\end{array}$

$\begin{array}{c}0.014+0.0050y=0.028-0.010y\\ 0.0050y+0.010y=0.028-0.014\\ y\left(0.015\right)=0.014\\ y=0.93\end{array}$

Calculate mole fractions of urea at equilibrium:

In beaker 1$\begin{array}{c}Molefraction=\frac{\text{0}\text{.0050}\text{mol}}{\text{0}\text{.0050}\text{mol+2}\text{.8}\text{mol-0}\text{.93}\text{mol}}\\ \text{=2}\text{.7}\times {\text{10}}^{\text{-3}}\end{array}$

In beaker 2$\begin{array}{c}Molefraction=\frac{\text{0}\text{.010}\text{mol}}{\text{0}\text{.010}\text{mol+2}\text{.8}\text{mol+0}\text{.93}\text{mol}}\\ \text{= 2}\text{.7}\times {\text{10}}^{\text{-3}}\end{array}$

Conclusion

The mole fraction of urea present in the two given solutions at equilibrium was calculated.