**If the total pressure of the mixture is 745mmHg, calculate the partial pressure acting on the krypton in that given mixture.**

This question aims to find the **partial pressure** exerted by an individual component of a **gaseous mixture**.

The basic concept behind this article on **Dalton’s Law of Partial Pressure** states that the **total pressure** that is exerted by a **mixture of gases** is the **accumulative sum** of **individual pressures** of **individual gas elements** that make up the mixture. It is represented as follows:

\[P_{Total}=P_{Gas1}+P_{Gas2}+P_{Gas3}+\ ……\]

It can also be expressed in terms of the **number of moles** or **mole fraction**:

\[P_{Gas1}=X_{Gas1}{\times P}_{Total}\]

Here $X_{Gas1}$ is the **Mole Fraction** for **Gas 1** which is represented as follows in terms of **number of moles** $n$:

\[X_{Gas1}\ =\frac{Number\ of\ moles\ of\ Gas1}{Sum\ of\ Number\ of\ moles\ of\ all\ Gases\ in\ the\ mixture}=\frac{n_{Gas1}}{n_{Gas1}+n_{Gas2}+n_{Gas3}+…..}\]

## Expert Answer

Given that:

**Percentage of Nitrogen Gas in the gaseous mixture** $N_2=75.2%$

**Percentage of Krypton Gas in the gaseous mixture** $Kr=24.8%$

**Total Pressure of the Gas Mixture** $P_{Total}=745\ mmHg$

**Molar Mass** of $N_2=28.013\dfrac{g}{mol}$

**Molar Mass** of $Kr=83.798\dfrac{g}{mol}$

We know that percentage of a gaseous component in a gas mixture represents the mass of the individual gas in **grams** $g$ per $100g$ of that particular gas mixture. Hence:

\[75.2\% \ of\ N_2=75.2g\ of\ N_2\]

\[24.8\% \ of\ Kr=24.8g\ of\ Kr\]

First, we will convert the given masses of individual gasses into the **number of moles** using **molar mass**.

We know that:

\[Number\ of\ Moles=\frac{Given\ Mass}{Molar\ Mass}\]

\[n=\frac{m}{M}\]

So, by using the above formula:

For **Nitrogen Gas** $N_2$:

\[n_{N_2}=\frac{75.2g}{28.013\dfrac{g}{mol}}\]

\[n_{N_2}=2.684mol\]

For **Krypton Gas** $Kr$:

\[n_{Kr}=\frac{24.8g}{83.798\dfrac{g}{mol}}\]

\[n_{Kr}=0.296mol\]

Now we will use the **Mole Fraction formula** for **Krypton Gas** as follows:

\[X_{Kr}=\frac{n_{Kr}}{n_{Kr}+{\ n}_{N_2}}\]

\[X_{Kr}=\frac{0.296mol}{0.296mol+2.684mol}\]

\[X_{Kr}=0.0993\]

To calculate the **Partial Pressure of Krypton** $Kr$, we will use **Dalton’s Law of Partial Pressure** in terms of **Mole Fraction** as follows:

\[P_{Kr}=X_{Kr}{\times P}_{Total}\]

Substituting the given and calculated values in the above equation:

\[P_{Kr}=0.0993\times745mmHg\]

\[Partial\ Pressure\ of\ Krypton\ Gas\ P_{Kr}=74.0mmHg\]

## Numerical Result

$24.8$ of Krypton Gas $(Kr)$ in a **gaseous mixture** having a **total pressure** of $745mmHg$ will exert an individual **partial pressure** of $74 mmHg$.

\[Partial\ Pressure\ of\ Krypton\ Gas\ P_{Kr}=74.0mmHg \]

## Example

A **gaseous mixture** comprising oxygen $21%$ and **Nitrogen** $79%$ exerts a **total pressure** of $750mmHg$. Calculate the **partial pressure** exerted by **Oxygen**.

**Solution**

**Percentage of Oxygen Gas in the gaseous mixture** $O_2=21%$

**Percentage of Nitrogen Gas in the gaseous mixture** $N_2=79%$

**Total Pressure of the Gas Mixture** $P_{Total}=750mmHg$

**Molar Mass** of $O_2=32\dfrac{g}{mol}$

**Molar Mass** of $N_2=28.013\dfrac{g}{mol}$

We know that:

\[21\%\ of\ O_2=21g\ of\ N_2\]

\[79\%\ of\ N_2=79g\ of\ Kr\]

We will convert the given masses of individual gasses into the **number of moles** using **molar mass**.

For **Oxygen Gas** $O_2$:

\[n_{O_2}=\frac{21g}{32\dfrac{g}{mol}}\]

\[n_{O_2}=0.656mol\]

For **Nitrogen Gas** $N_2$:

\[n_{N_2}=\frac{79g}{28.013\dfrac{g}{mol}}\]

\[n_{N_2}\ =\ 2.82mol\]

In order to calculate the **Partial Pressure of Oxygen** $O_2$, we will use the **Dalton’s Law of Partial Pressure** in terms of **Mole Fraction** as follows:

\[P_{O_2}=X_{O_2}{\times P}_{Total}\]

\[P_{O_2}=\frac{n_{O_2}}{n_{N_2}+\ n_{O_2}}{\times P}_{Total} \]

\[P_{O_2}=\frac{0.656mol}{0.656\ mol+2.82\ mol} \times750mmHg\]

\[Partial\ Pressure\ of\ Oxygen\ Gas\ P_{O_2}=141.54mmHg\]