Question

Which of the following is the correct realtion between Molarity, Normality and n-factor $\left(n_f\right)$?

• A. $\text{Molarity}=\text{Normality}\times \text{n-factor}$
• B. $\text{Normality}\times \text{Molarity}=\text{n-factor}$
• C. $\text{Normality}=\text{Molarity}\times \text{n-factor}$
• D. None of the above

Answer 1

The correct option is C $\text{Normality}=\text{Molarity}\times \text{n-factor}$

$\text{Molarity}=\frac{\text{Number of moles of solute}}{\text{Volume of solution (L)}}.......\left(A\right)$

Since,
$\text{Normality}=\frac{\text{Number of g-equivalents of solute}}{\text{Volume of solution (L)}}\text{Number of g-equivalents of solute}=\frac{\text{Weight of solute}}{\text{Equivalent weight of solute}}\text{Equivalent weight of solute}=\frac{\text{Molecular Mass of solute}}{\text{n-factor}}$

Putting the values we get,

$\text{Normality}=\frac{\text{Weight of solute}\times \text{n-factor}}{\text{Molecular Mass of solute}\times \text{Volume of solution (L)}}\text{Normality}=\frac{\text{Number of moles of solute}}{\text{Volume of solution (L)}}\times \text{n-factor}$
From Equation (A)

$\text{Normality}=\text{Molarity}\times \text{n-factor}$