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 $55.5 M$.