Mass and Mole Fractions

  • Apply the basic concepts of non-reacting mixtures in general and air-vapour mixtures in particular system.

  • Perform calculations to determine the mole fractions of gases within and gas mixture and relate mole fraction to the partial pressure of a gas within a gas mixture.

  • Mass fraction, \(m_f\): the ratio of the mass of a component to the mass of the mixture

  • Mole fraction, y: the ratio of the mole number of a component to the mole number of the mixture.

Solved Example:

76-1-01

What is the mole fraction of Ascorbic acid, C$_6$H$_8$O$_6$, in a solution containing 80.5 g of Ascorbic acid and 210 g of water?

Solution:
\[\mathrm{Mole\ Fraction} = \dfrac{\mathrm{No.\ of\ moles\ of\ element}}{\mathrm{Total\ No.\ of\ moles}} =\dfrac {80.5}{\left( 12\times 6\right) +\left( 1\times 8\right) +\left( 16\times 6\right) } =0.457\] \[\mathrm{Number\ of\ moles\ of\ H_2O} =\dfrac {210}{18} =11.667\] \[\mathrm{Mole\ fraction}=\dfrac {0.457}{0.457+11.667}=0.0377\]

Correct Answer: A

Solved Example:

76-1-02

A mixture of 1.00 mole of O$_2$ and an unknown number of moles of N$_2$ is maintained under standard conditions. If 3.00 moles of O$_2$ are added to the mixture, maintaining the same conditions, the volume doubles. How many moles of N$_2$ are present?

Solution:
Initial number of moles of O$_{2}$=1= n$_{A1}$
Initial number of moles of N$_{2}$=x= n$_{B}$
Final number of moles of O$_{2}$= 1 + 3 = 4= n$_{A2}$
From Avagadro's law, \[\dfrac {V_{1}}{n_{1}}=\dfrac {V_{2}}{n_{2}}\] \[n_{1}=n_{A1}+ n_{B}\] \[n_{1}=x+1\] Where x is the number of moles of N$_{2}$ \[n_{2}=n_{A2}+n_{B} =x+4\] \[V_{2}=2V_{1}\] \begin{align*} \dfrac{V_{1}}{x+1} &=\dfrac {V_{2}}{x+4}\\ \dfrac{V_{1}}{x+1} &=\dfrac {V_{2}}{x+4}\\ \dfrac{V_{1}}{x+1} &=\dfrac {2V_{1}}{x+4}\\ x+4 &=2\times \left( x+1\right) \\ x+4 &=2x+2\\ x &=2 \end{align*}

Correct Answer: B