Question

The equivalent weight of HCl in the given reaction is:
${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }\quad +\quad 14HCl\quad \longrightarrow \quad 3KCl\quad +\quad 2Cr{ Cl }_{ 3 }\quad +\quad 3{ Cl }_{ 2 }\quad +\quad 7{ H }_{ 2 }O$
A. 16.25
B. 36.5
C. 73
D. 85.1

Answer 1

Equivalent is the quantity that combines with or replaces 1 g of hydrogen or 8 g of oxygen. Molecular weight is the sum of the weight of the individual atoms of the molecule.

Complete step by step answer:
Equivalent weight can be defined as the one equivalent which is the mass of a given substance which combines with or replaces a fixed quantity of another substance. It is the mass of a substance that combines with or is chemically equivalent to one gram of hydrogen or eight grams of oxygen.

In the given reaction HCl is getting oxidized. The following redox reaction takes place.
${ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }\quad +\quad 14{ H }^{ + }\quad +\quad 6{ e }^{ - }\quad \longrightarrow \quad { 2Cr }^{ 3+ }\quad +\quad 7{ H }_{ 2 }O$

Here, we can see that 14 moles of HCl is losing 6 moles of ${e}^{-}$. This means that 1 mole of HCl is losing $\dfrac {6} {14}$ moles of ${e}^{-}$.
Therefore, n-factor = $\dfrac {6} {14}$
Now equivalent mass is given as the molecular weight divided by the n-factor.

$Equivalent\quad mass\quad =\quad \dfrac { Molecular\quad Weight }{ n-factor }$ ----(1)

The molecular weight of HCl is 35.5 + 1 = 36.5
Substituting the values in the equation (1), we get,
$Equivalent\quad mass\quad =\quad \cfrac { 36.5 }{ \cfrac { 6 }{ 14 } }$
$\implies Equivalent\quad mass\quad =\quad \dfrac { 36.5\quad \times \quad 14 }{ 6 }$
$\implies Equivalent\quad mass\quad =\quad 85.1g$

Therefore, the equivalent mass of HCl is 85.1g. Hence, the correct option is option (D).

Note: Students tend to understand the molecular weight and the equivalent weight of a compound as equal. But they are different from each other. Molecular weight is the sum of the weights of the individual atoms in the molecule whereas, the equivalent weight is molecular weight divided by the n-factor.