Question

What is the equivalent weight of glucose in the reaction given below?
${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}{\text{ + 6}}{{\text{O}}_{\text{2}}} \to {\text{6C}}{{\text{O}}_{\text{2}}}{\text{ + 6}}{{\text{H}}_{\text{2}}}{\text{O}}$
(M is the molecular weight)
(A) $\dfrac{{\text{M}}}{{\text{4}}}$
(B) $\dfrac{{\text{M}}}{{12}}$
(C) $\dfrac{{\text{M}}}{{{\text{24}}}}$
(D) $\dfrac{{\text{M}}}{{{\text{48}}}}$

Answer 1

In the above question, we are provided with an equation and asked to find the equivalent weight of glucose. Since, we are asked to take molecular weight as M and hence the only thing we have to do is to find the change in valence electrons of C while the reaction is taking place.
Equivalent weight = $\dfrac{{{\text{molecular mass}}}}{{{\text{change in valence electrons}}}}$
Oxidation state = charge on the molecule - $\sum\limits_{{\text{i = 0}}}^{\text{k}} {{{\text{n}}_{\text{i}}}{{\text{c}}_{\text{i}}}}$
Where ${{\text{n}}_{\text{i}}}$ = number of ${{\text{i}}^{{\text{th}}}}$ element
${{\text{c}}_{\text{i}}}$ = charge on ${{\text{i}}^{{\text{th}}}}$ element.

Complete step by step solution:
The equivalent weight of an element is defined as the ratio of atomic mass and the oxidation state.
Let us now find out the oxidation state:
Oxidation state of C in ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{{\text{O}}_{\text{6}}}$ :
Let the oxidation state of C be x.
x= charge on the molecule ${\text{ - }}\sum\limits_{{\text{i = 0}}}^{\text{k}} {{{\text{n}}_{\text{i}}}{{\text{c}}_{\text{i}}}}$
which can be written as-
x= charge on the molecule ${\text{ - 12 \times }}$ charge on H ${\text{ - 6 \times }}$ charge on O.
Since charge on H atom is ${\text{ + 1}}$ and O atom is ${\text{ - 2}}$ . And charge on the molecule is 0.
Hence, x = ${\text{0 - 12 \times 1 - 6 \times ( - 2) = 0}}$
Therefore, here valence electrons are 0.
Oxidation state of C in ${\text{C}}{{\text{O}}_{\text{2}}}$ :
Let the oxidation state of C be x.
x= charge on the molecule ${\text{ - }}\sum\limits_{{\text{i = 0}}}^{\text{k}} {{{\text{n}}_{\text{i}}}{{\text{c}}_{\text{i}}}}$
which can be written as-
x= charge on the molecule ${\text{ - 2 \times }}$ charge on O.
Since charge of O atom is ${\text{ - 2}}$ and charge on the molecule is 0.
Hence, x = ${\text{0 - 2 \times ( - 2) = 4}}$
Since, 6 molecules of carbon dioxide is present hence, valence electrons is equal to ${\text{6 \times 4 = 24}}$
Change in valence electrons = ${\text{24 - 0 = 24}}$ .
As we know that Equivalent weight = $\dfrac{{{\text{molecular mass}}}}{{{\text{change in valence electrons}}}}$
So by substituting the values, we get:
Equivalent weight= $\dfrac{{\text{M}}}{{{\text{24}}}}$
Hence, the correct option is option C.

Note:
Basically we use the idea of equivalent mass is to compare chemically different elements. Atoms combine with each other to form chemical compounds which implies that the elements are always present in definite proportions by mass and this property can be used to make chemically different molecules with different mass, equal.