Question

A solution contains $1.2046\times {{10}^{24}}$hydrochloric acid molecules in one $\text{d}{{\text{m}}^{\text{3}}}$of the solutions. The strength of the solution is:
A) 6N
B) 2N
C) 4N
D) 8N

Answer 1

The Avogadro's number $\text{NA}$ is a proportionality constant. It relates the number of particles in a sample with the amount of substance in the sample. The strength of the solution is defined as the number of moles per unit volume.

Complete answer:
We know that Avogadro's number relates to the number of particles in the sample with the number of moles of the sample. In other words, one mole of a substance contains the $\text{6}\text{.023 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}$particles.
We are given with,
The $1\text{ d}{{\text{m}}^{\text{3}}}$ Hydrochloric acid solution contains $1.2046\times {{10}^{24}}$ a molecule.
Since $1\text{ mole=6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{particle}$
Then,$\text{x mole=1}\text{.2046}\times \text{1}{{\text{0}}^{\text{24}}}\text{ molecule of HCl }$
Therefore, $\text{No}\text{.of mole of HCl =}\dfrac{1.2046\times {{10}^{24}}\text{particles}}{\text{6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{particle}}$
Therefore,$\text{No}\text{.of mole of HCl =}2\text{ moles}$
Now, we have to find the strength of the solution.
The normality of a solution is described as the number of gram equivalents of solute per litre of solution.
$\text{Normality=}\dfrac{\text{No}\text{. of gram equivalent of solute}}{\text{Volume of solution in litre}}$.
Since, $\text{No}\text{. of gram equivalent of solute}=\dfrac{\text{weight of solute in gm}}{\text{Equivalent weight}}$
But we know that,$\text{Equivalent weight=}\dfrac{\text{molecular weight of solute}}{\text{Valence factor}}$
Where the valence factor (n) depends on the solute.
On combining the equation of the number of gram equivalent, equivalent weight with the formula of normality. We get,
$\text{Normality=}\dfrac{\dfrac{\text{weight of solute }}{\text{molecular weight}}}{\text{Volume of solution in litre}}\times \text{Valence factor}$
But we know that, $\text{No of moles=}\dfrac{\text{weight of solute }}{\text{molecular weight}}$
Therefore, the equation of normality can be written as,
$\text{Normality=}\dfrac{\text{No}\text{.of moles}}{\text{Volume of solution in litre}}\times \text{Valence factor}$
The valence factor or n-factor for a$\text{ HCl }$is equal to the basicity of the acid. Since $\text{ HCl }$loses its one proton. The valence factor or n-factor for $\text{ HCl }$is 1.
We also know that $\text{1d}{{\text{m}}^{\text{3}}}\text{=1000mL=1L}$
Let's substitute these values in the formula of normality.
$\text{Normality=}\dfrac{2\text{ mol}}{1\text{L}}\times \text{1}$
$\text{Normality=}2\text{ mol }{{\text{L}}^{\text{-1}}}\text{ or 2N}$
Thus, the strength of the solution is $\text{2N}$

Hence, (B) is the correct option.

Additional information:
Molarity is also used to calculate the strength of the solution. It is defined as the number of moles per unit volume in litre. The normality and molarity are related to each other.
$\text{Normality=}\dfrac{\text{No}\text{.of moles}}{\text{Volume of solution in liter}}\times \text{Valence factor}$
We know that,
$\text{Molarity=}\dfrac{\text{No}\text{.of moles}}{\text{Volume of solution in liter}}$.
Thus on substituting in the equation of normality,
$\text{Normality=Molarity}\times \text{Valence factor}$
Or can be written as,$\text{N=nM}$

Note:
To determine the normality of solution primarily it is needed to find the reactive species in the solution. In a chemical reaction, a gram equivalent weight is determined as the amount of ion which reacts. This depends on the reaction. Thus, the gram equivalent is not a consistent amount.