Question

The equivalent weight of HCl in the given reaction is:
${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }{ +14HCl\rightarrow 2KCl+2CrCl }_{ 3 }{ +3Cl }_{ 2 }{ +H }_{ 2 }{ O }$
a.) ${ 16.25 }$
b.) ${ 36.5 }$
c.) ${ 73 }$
d.) ${ 85.1 }$

Answer 1

Hint: The equivalent weight of a compound is its molecular weight divided by its valence (number of electrons gained or lost by one molecule or ion of the substance in the reaction).

Complete answer:
The following reaction is given:
${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }{ +14HCl\rightarrow 2KCl+2CrCl }_{ 3 }{ +3Cl }_{ 2 }{ +H }_{ 2 }{ O }$

In this reaction, HCl is oxidized to ${ Cl }_{ 2 }$.
So, the change in oxidation number = ${ 0 - (-1) = 1 }$
Now, we have to calculate n-factor.
n-factor represents the number of ${ H }^{ + }$ ions an acid can give or the number of ${ OH }^{ - }$ ions a base can produce.
n-factor = $\dfrac { 6 }{ 14 }$
As we know,

The molecular mass of HCl = ${ 36.5gmol }^{ -1 }$
Equivalent weight = $\dfrac { molecular\quad mass\quad of\quad HCl }{ n-factor }$
By putting the values in the above formula, we get
Equivalent weight = $\dfrac { 36.5\times 14 }{ 6 }$
Equivalent weight = ${ 85.1 }$

Hence, the equivalent weight of HCl in the given reaction is ${ 85.1 }$.
So, the correct option is D.

Additional Information:
Uses of potassium dichromate:
It is utilized in photographic screen printing.
It is a typical reagent utilized in analytical science for classical "wet tests".
It is utilized to recolor specific kinds of woods to deliver profound, rich brown colors.
It is utilized to clean dish sets and as a drawing material.

Note: The possibility to make a mistake is that you may choose option A. But the n-factor is $\dfrac { 6 }{ 14 }$ not 2 as the change in the oxidation state is ${ +6 }$.