# How many liters of a 0.0550m KCl solution contain 0.163 moles of KCl?

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:

$Molarity M = \frac{n}{V} = \frac{N}{N_AV}$

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 $55.5 M$.