Question

The equivalent weight of $ C{H_4} $ in the reaction $ C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O $ is [M=molecular weight]
(A) $ \dfrac{M}{4} $
(B) $ \dfrac{M}{8} $
(C) $ \dfrac{M}{{12}} $
(D) $ \dfrac{M}{{16}} $

Answer 1

Hint :Here, we need to find out the equivalent weight of $ C{H_4} $ . Equivalent weight of any atom or any molecule can be calculated by dividing its atomic or molecular weight by the valency of that atom or molecule. Another name of equivalent weight is gram equivalent.

Complete Step By Step Answer:
Now, since we need to find out the equivalent weight of $ C{H_4} $ so we are going to use a formula, that is,
Equivalent Weight = $ \dfrac{M}{V} $ , where M stands for molecular weight and V stands for valency, which is basically the capacity of an atom to combine with other atoms.
Now, let’s calculate the equivalent mass of $ C{H_4} $ -
 $ \dfrac{{A{M_c}}}{V} + \dfrac{{A{M_H}}}{V} $ ,
where $ A{M_c} $ stands for the atomic mass of Carbon,
 $ A{M_H} $ stands for the atomic mass of Hydrogen,
 $ V $ stands for the valency.
So, we know the atomic mass of Carbon and Hydrogen is $ 12 $ and $ 1 $ , respectively and the valency for Carbon and Hydrogen in $ C{H_4} $ is $ 4 $ and $ 2 $ .
Now, we will put these values in the formula-
 $ \dfrac{{12}}{4} + \dfrac{1}{2} = 3 + 1 = 4 $
Now, molecular weight of $ C{H_4} = 12 + (4 \times 1) = 16u $
So, the equivalent mass of $ C{H_4} = \dfrac{M}{V} = \dfrac{{16}}{4} = \dfrac{M}{4} $
Therefore, the correct option is $ 1)\dfrac{M}{4} $ .

Note :
The molecular weight of any compound can be easily calculated by adding all the atomic weight and multiplying them by the number of atoms present with their corresponding atomic weights.