Chapter 10, Problem 39PS

A halothane-oxygen mixture (C₂HBrCIF₃ + O₂) can be used as an anesthetic. A tank containing such a mixture has the following partial pressures: P (halothane) = 170 mm Hg and P (O₂) = 570 mm Hg.

1. (a) What is the ratio of the number of moles of halothane to the number of moles of O₂?
2. (b) If the tank contains 160 g of O₂, what mass of C₂HBrCIF₃ is present?

Interpretation Introduction

Interpretation:

The ratio of number of moles of halothane vapour to oxygen gas and mass of ${\text{C}}_{\text{2}}{\text{HBrClF}}_{\text{3}}$ has to be given.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

• $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \end{array}$

  Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

Answer to Problem 39PS

The ratio of number of moles of halothane to the ratio of number of moles of oxygen is

  $\frac{\text{0}\text{.3}\text{mol}\text{halothane}}{\text{1}\text{mol}{\text{O}}_{\text{2}}}$

Explanation of Solution

Given:

  $\begin{array}{l}\text{p(halothane)}\text{=}\text{170}\text{mm}\text{Hg}\\ {\text{p(O}}_{\text{2}}\text{)}\text{=}\text{570}\text{mm}\text{Hg}\end{array}$

Total pressure can be found out by taking the sum of these two partial pressures.

  $\begin{array}{l}\text{Total}\text{pressure}\text{=}\text{570}\text{mm}\text{Hg}\text{+}\text{170}\text{mm}\text{Hg}\\ \text{=}\text{740}\text{mm}\text{Hg}\end{array}$

Now we can calculate the mole fraction by the equation

  $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ {\text{x}}_{\text{a}}\text{=}\frac{{\text{p}}_{\text{a}}}{{\text{P}}_{\text{total}}}\\ \end{array}$

  $\begin{array}{l}{\text{x}}_{\text{halothane}}\text{=}\frac{\text{170}\text{mm}\text{Hg}}{\text{740}\text{mm}\text{Hg}}\\ \text{=}\text{0}\text{.229}\end{array}$

  $\begin{array}{l}{\text{x}}_{\text{halothane}}\text{=}\frac{\text{570}\text{mm}\text{Hg}}{\text{740}\text{mm}\text{Hg}}\\ \text{=}\text{0}\text{.770}\end{array}$

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

We know the mole fraction of both the given compounds and using this, ratio of number of moles can be found out:

  $\begin{array}{l}\frac{{\text{x}}_{\text{halothane}}}{{\text{x}}_{{\text{O}}_{\text{2}}}}\text{=}\frac{\frac{{\text{n}}_{\text{halothane}}}{{\text{n}}_{\text{total}}}}{\frac{{\text{n}}_{{\text{O}}_{\text{2}}}}{{\text{n}}_{\text{total}}}}\\ \frac{\text{0}\text{.229}}{\text{0}\text{.770}}\text{=}\frac{{\text{n}}_{\text{halothane}}}{{\text{n}}_{{\text{O}}_{\text{2}}}}\\ \frac{\text{0}\text{.297}}{\text{1}}\text{=}\frac{{\text{n}}_{\text{halothane}}}{{\text{n}}_{{\text{O}}_{\text{2}}}}\\ \frac{\text{0}\text{.3}}{\text{1}}\text{=}\frac{{\text{n}}_{\text{halothane}}}{{\text{n}}_{{\text{O}}_{\text{2}}}}\end{array}$

The ratio of number of moles of halothane to the ratio of number of moles of oxygen is

     $\frac{\text{0}\text{.3}\text{mol}\text{halothane}}{\text{1}\text{mol}{\text{O}}_{\text{2}}}$

Interpretation Introduction

Interpretation:

The ratio of number of moles of halothane vapour to oxygen gas and mass of ${\text{C}}_{\text{2}}{\text{HBrClF}}_{\text{3}}$ has to be given.

Concept Introduction:

• Partial pressure: The pressure of each gas in a mixture of gases is the partial pressure.
• Dalton’s law of partial pressure: The total pressure of gases in a mixture of gases is the sum of the pressure of each gas in the mixture
• Mole fraction: Quantity which defines the number of moles of a substance in a mixture divided by the total number of moles of all substances present.

  ${\text{x}}_{\text{a}}\text{=}\frac{{\text{n}}_{\text{a}}}{{\text{n}}_{\text{total}}}$

• $\begin{array}{l}{\text{p}}_{\text{a}}\text{=}{\text{x}}_{\text{a}}\text{×}{\text{P}}_{\text{total}}\\ \end{array}$

  Partial pressure of a gas in the mixture of gases is the product of mole fraction of the gas and the total pressure.

Answer to Problem 39PS

The mass of halothane is found to be ${\text{3×10}}^{\text{2}}\text{g}$

Explanation of Solution

The mass of halothane if the tank contain $\text{160}\text{g}\text{of}{\text{O}}_{\text{2}}$

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{of}{\text{O}}_{\text{2}}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{=}\frac{\text{160}\text{g}}{\text{32}\text{g}}\\ \text{=}\text{5}\text{mol}\\ \end{array}$

The ratio of number of moles of halothane to number of moles of oxygen is calculated to be

  $\frac{\text{0}\text{.3}\text{mol}\text{halothane}}{\text{1}\text{mol}{\text{O}}_{\text{2}}}$

So if 5 moles of ${\text{O}}_{\text{2}}$ is present, number of moles of halothane will be $\text{0}\text{.3×5=}\text{1}\text{.5}\text{mol}$

  $\begin{array}{l}\text{Number}\text{of}\text{moles}\text{=}\frac{\text{mass}}{\text{molar}\text{mass}}\\ \text{mass}\text{of}\text{halothane}\text{=}\text{Number}\text{of}\text{moles×}\text{molar}\text{mass}\\ \text{=}\text{1}\text{.5}\text{×}\text{197}\text{.38}\text{g}\\ \text{ =}\text{2}\text{.96}{\text{×10}}^{\text{2}}\text{g}\\ \text{=}{\text{3×10}}^{\text{2}}\text{g}\end{array}$