The questions aim to find the moles of $KCl$ from the $KCl$ solution.

In this problem, we will use the concept of molar solution and molarity to calculate the volume of $KCl$. **Molarity (M)** or **Molar Concentration of a Solution** is defined as the concentration of a substance, usually a solute, in a particular volume of solution. Molarity is calculated in terms of the number of moles of a solute per liters of a solution.

**Molarity** is defined by Unit $M$ or $\dfrac{mol}{L}$. A $1M$ solution is said to have a concentration of **“one molar”**

\[ Molarity M = \frac{n}{V} \]

Where:

$ n = \text{Number of Moles}$

$ V = \text{Volume of Solution in Liters}$

As:

\[ n = \frac{N}{N_A} \]

Where:

$ N = \text{Number of Particle of given solute in Volume $V$ of solution} $

$ N_A = Avogadros Number = 6.022 \times {10}^{23} $

So:

\[ Molarity M = \frac{n}{V} = \frac{N}{N_AV} \]

**Expert Answer**

Number of Moles of KCl $ n = 0.163 mol$

**Molarity** of $KCl$ Solution $M = 0.0550 M$ or $M = 0.0550 \dfrac{mol}{L}$

We need to find the **volume** of $KCl$ Solution $V=?$

By using the concept of molarity, we know that:

\[ Molarity M = \frac{n}{V} \]

Substituting the values of $n = 0.163 mol$ and $M = 0.0550 \dfrac{mol}{L}$ in above equation, we get:

\[ 0.0550 \frac{mol}{L} = \frac{0.163mol}{V} \]

\[ V = \frac{0.163mol}{0.0550 \dfrac{mol}{L}} \]

By canceling the units of mol and solving the equation, we get:

\[ V = 2.964L \]

## Numerical Results

Volume $V$ of $KCl$ in a solution of $0.0550 M$ and containing $0.163 moles$ $ = 2.964 L$

## Example

Calculate the **Molarity** of Pure Water, which is having a volume of **$1000 ml$** if its **density** is assumed to be $1 \dfrac{g}{ml}$

Given as:

**Density of Water** $\rho = 1 \dfrac{g}{ml}$

Volume of Water $V = 1000 ml = 1 L$

**Molarity** $M$ of Pure Water $=?$

We know that:

\[ Mass\ of\ Water\ m = Volume \times Density = V \times \rho\]

\[ Mass\ of\ Water\ m = 1000 ml \times 1 \frac{g}{ml} = 1000g\]

**Molar Mass of Pure Water** $H_2O$ is

\[ M = 1g \times 2 + 16g \times 1 = (2+16) g = 18g \]

**Number of Moles** $n$ of Pure water $H_2O$

\[ n = \frac{Given\ Mass\ m}{Molecular\ Mass\ M} = \frac{1000g}{18g} = 55.5\ mol \]

By using the concept of **Molarity**, know that:

\[ Molarity\ M = \frac{n}{V} \]

By substituting the values of $n$ and $V$

\[ Molarity\ M = \frac{55.5\ mol}{1L} = 55.5 \frac{mol}{L} \]

Hence, **molarity** of $1000 ml$ of Pure water is $55.5 M$.