Question

If true enter 1, else enter 0.
Normality $=$ Molarity $\times$ Valence factor
A.1
B.0

Answer 1

The term ‘normality of a solution’ is used to refer to the number of gram equivalents of the solute dissolved per litre or ${\text{d}}{{\text{m}}^{\text{3}}}$ of the solution. It is represented by the letter N.
The term ‘molarity of a solution’ is used to refer to the number of moles of the solute dissolved per litre or ${\text{d}}{{\text{m}}^{\text{3}}}$ of the solution. It is represented by the letter M.

Complete step by step answer:
Mathematically, normality of a solution ‘N’ is equal to the number of gram equivalent of solute divided by the volume of the solution in litres.
If ‘w’ gram of the solute is present in V ${\text{c}}{{\text{m}}^{\text{3}}}$ of a given solution, then:
${\text{N}} = \dfrac{{\text{w}}}{{{\text{Eq}}{\text{.mass}}\left( {{\text{solute}}} \right)}} \times \dfrac{{1000}}{{\text{V}}}$
Mathematically, molarity of a solution ‘M’ is equal to the number of moles of solute divided by the volume of the solution in litres.
If ‘w’ gram of the solute is present in V ${\text{c}}{{\text{m}}^{\text{3}}}$ of a given solution, then:
${\text{M}} = \dfrac{{\text{w}}}{{{\text{Mol}}{\text{.mass}}\left( {{\text{solute}}} \right)}} \times \dfrac{{1000}}{{\text{V}}}$
Relationship between normality and molarity of a solution:
Suppose a solution is ‘x’ molar and molecular mass of the solute is M and its equivalent mass is E. Then,
Molarity of the solution ${\text{ = xmol}}{{\text{L}}^{{\text{ - 1}}}}$
If we put the value of molecular mass, the molarity is obtained in gram per litre.
Molarity of the solution ${\text{ = x}} \times {\text{Mg}}{{\text{L}}^{{\text{ - 1}}}}$
Therefore,
Molarity of the solution ${\text{ = }}\dfrac{{{\text{x}} \times {\text{M}}}}{{\text{E}}}{\text{geq}}{{\text{L}}^{{\text{ - 1}}}}$
This is equal to normality of the solution. So, the normality of the solution ${\text{ = }}\dfrac{{{\text{x}} \times {\text{M}}}}{{\text{E}}}$ .
Thus, in general, normality of a solution $=$ Molarity $\times$ $\dfrac{{{\text{Mol}}{\text{.mass}}}}{{{\text{Eq}}{\text{.mass}}}}$ .
But, the equivalent mass of a molecule is equal to the molecular mass of the molecule divided by the valency of the molecule.
So, normality of a solution $=$ Molarity $\times$ $\dfrac{{{\text{Mol}}{\text{.mass}} \times {\text{Valency}}}}{{{\text{Mol}}{\text{.mass}}}}$
This gives -
Normality $=$ Molarity $\times$ Valence factor
So, the given statement is true

So, the correct answer is A.

Note: For an acid, molecular mass divided by the equivalent mass of the acid is equal to basicity of the acid and so, normality of an acid $=$ Molarity $\times$ Basicity
For a base, molecular mass divided by the equivalent mass of the base is equal to acidity of the base and so, normality of a base $=$ Molarity $\times$ Acidity