This question is from the **chemistry** domain and aims to explain the basic **concepts** related to the **periodic** table, **Molar** masses, the ratio of **effusion** rates, and **Graham** law to determine the **effusion** rate.

In the **periodic** table, the **chemicals** are displayed in a **tabular** manner. The periodic table sometimes is also named the **periodic** table of the chemical **elements.** Periodic is considered the **icon** of chemistry and is widely used in **physics, chemistry,** and other **sciences.** periodic law **states** that the properties of the **chemical** elements show a systematic **dependence** on their **atomic** numbers. The rows in the **periodic** table are called the **period** and the columns are called the **groups.** Elements in the same groups represent the same **chemical** properties. There is a total of $18$ groups and $7$ periods in the **periodic** table.

**Mass** in grams of one **mole** of the element in the **molar** mass of the substance. The amount of entities e.g **molecules, ions,** and **atoms** present in a **substance** is defined as a **mole.** A mole of any **substance** is given by:

\[ 6.022 \times 10^{22} \]

The rate of **effusion** of a gaseous **essence** is inversely **proportional** to the square **root** of its molar mass. This **relationship*** is *directed to **Graham’s law,** after the Scottish chemist Thomas Graham (1805–1869). The proportion of the **effusion** rates of two gases is the square root of the inverse ratio of their **molar** masses*: *

\[ \dfrac{rate_{Ar}}{rate_{Kr}} = \sqrt{ \dfrac{M_{Kr}}{M_{Ar}}} \]

## Expert Answer

**Ar** is a chemical element that represents **Argon** gas. Argon has the atomic number $18$ and lies in the **group** $18$ of the periodic table. **Argon** is a **noble** gas and has a **molecular** mass of $39.948$.

**Kr** is a chemical element that represents **Krypton.** Krypton has the **atomic** number 36 and lies in the group $18$ of the **periodic** table. **Krypton** is an odorless, **tasteless,** and colorless noble **gas** and has a **molecular** mass of $83.789$.

**Molar** mass of argon is $M_{Ar} =39.948 g/mol$

**Molar** mass of Krypton is $M_{Kr} =83.798 g/mol$

To determine the ratio of **effusion** rates, **substitute** the molar masses in **Graham’s** law:

\[ \dfrac{rate_{Ar}}{rate_{Kr} }= \sqrt{\dfrac{M_{Kr}}{M_{Ar}}} \]

\[ \dfrac{rate_{Ar}} {rate_{Kr}} = \sqrt{ \dfrac{83.798} {39.748}} \]

\[ \dfrac{rate_{Ar}} {rate_{Kr}}= 1.4483 \]

## Numerical Answer

The ratio of **effusion** rates for $Ar$ and $Kr$ is $1.4483$.

## Example

**Unknown** gas has a rate of effusion of $9.20 mL/min$. The rate of **effusion** of pure nitrogen (N_2) gas under **identical** conditions is $14.64 mL/min$. Using **Graham’s** law, identify the **unknown** gas.

\[ \dfrac{rate_{X}}{rate_{N_2}} = \sqrt{ \dfrac{M_{N_2}} {M_{X}} } \]

We have to **identify** the $M_X$ to, For this, we are taking **squares** on both sides.

\[ \left( \dfrac{rate_{X}}{rate_{N_2}} \right)^2 = \dfrac{M_{N_2}} {M_{X}} \]

**Plugging** the values and solving for $M_X$:

\[ \left( \dfrac{9.20 \space mL/min}{14.65 \space mL/min} \right)^2 = \dfrac{28.0 \space g/mol} {M_{X}} \]

\[ M_x = \dfrac{28.0 \space g/mol} {\left( \dfrac{9.20 \space mL/min}{14.65 \space mL/min} \right)^2} \]

\[M_X = 71.0 \space g/mol \]

$71.0 \space g/mol$ is the molar **mass** of the unknown gas.