Question

Determine equivalent weight of carbon in the following:
(a) ${\rm{C}} + {{\rm{O}}_{\rm{2}}} \to {\rm{C}}{{\rm{O}}_{\rm{2}}}$
(b)${\rm{C}} + {\rm{2}}{{\rm{H}}_{\rm{2}}} \to {\rm{C}}{{\rm{H}}_4}$
(c)${\rm{2C}} + {{\rm{H}}_{\rm{2}}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}$
(d)${\rm{2C}} + {\rm{2}}{{\rm{H}}_{\rm{2}}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}$
Which of the following options is correct?
(A) (a) $16\;{\rm{g}}\;{{\rm{O}}_{\rm{2}}} \equiv {\rm{3}}\;{\rm{g}}\;{\rm{C}}$
(b) ${\rm{1}}\;{\rm{g}}\;{\rm{H}} \equiv {\rm{3}}\;{\rm{g}}\;{\rm{C}}$
(c)${\rm{1}}\;{\rm{g}}\;{{\rm{H}}_{\rm{2}}} \equiv 12\;{\rm{g}}\;{\rm{C}}$
(d)${\rm{1}}\;{\rm{g}}\;{{\rm{H}}_{\rm{2}}} \equiv 6\;{\rm{g}}\;{{\rm{H}}_2}$

(B) (a) ${\rm{4}}\;{\rm{g}}\;{{\rm{O}}_{\rm{2}}} \equiv 1\;{\rm{g}}\;{\rm{C}}$
(b) ${\rm{1}}\;{\rm{g}}\;{\rm{H}} \equiv {\rm{3}}\;{\rm{g}}\;{\rm{C}}$
(c)${\rm{1}}\;{\rm{g}}\;{{\rm{H}}_{\rm{2}}} \equiv 12\;{\rm{g}}\;{\rm{C}}$
(d)${\rm{1}}\;{\rm{g}}\;{{\rm{H}}_{\rm{2}}} \equiv 3\;{\rm{g}}\;{{\rm{H}}_2}$

(C) (a) ${\rm{6}}\;{\rm{g}}\;{{\rm{O}}_{\rm{2}}} \equiv {\rm{3}}\;{\rm{g}}\;{\rm{C}}$
(b)${\rm{1}}\;{\rm{g}}\;{\rm{H}} \equiv 2\;{\rm{g}}\;{\rm{C}}$
(c)${\rm{2}}\;{\rm{g}}\;{{\rm{H}}_{\rm{2}}} \equiv 10\;{\rm{g}}\;{\rm{C}}$
(d)${\rm{1}}\;{\rm{g}}\;{{\rm{H}}_{\rm{2}}} \equiv 6\;{\rm{g}}\;{{\rm{H}}_2}$

(D) None of these

Answer 1

As we know that, the equivalent weight of any compound is the weight of compound which undergoes dissociation or association with the other fixed mass of compound. Any atom in the compound has its own valency and it changes in the reactants and products.

Step by step answer: The equivalent weight is calculated by the ratio of molar mass of the atom or molecule by change in the oxidation state as-
${\rm{Equivalent}}\,{\rm{weight = }}\dfrac{{{\rm{molar}}\;{\rm{mass}}}}{{{\rm{change}}\,{\rm{in}}\,{\rm{oxidation}}\,{\rm{state}}}}$
In the reaction (a) ${\rm{C}} + {{\rm{O}}_{\rm{2}}} \to {\rm{C}}{{\rm{O}}_{\rm{2}}}$
The molar mass of carbon is ${\rm{12}}\;{\rm{g}}$and the change in oxidation state of carbon from carbon to carbon dioxide is-
Oxidation state of carbon in ${\rm{C}} = {\rm{0}}$
Oxidation state of carbon in ${\rm{C}}{{\rm{O}}_2} = {\rm{ + 4}}$
SO, the change in oxidation state$= {\rm{4}}$

$${\rm{Equivalent}}\,{\rm{weight\;of \;C}} = \dfrac{{1{\rm{2}}\;{\rm{g}}}}{4}\\\ {\rm{Equivalent}}\,{\rm{weight\;of\;C}} = {\rm{3}}\;{\rm{g}}$$

Because the options are also given for oxygen so we have to calculate equivalent weight of oxygen as well-
Oxidation state of oxygen in ${{\rm{O}}_2} = {\rm{0}}$
Oxidation state of oxygen in ${\rm{C}}{{\rm{O}}_2} = - {\rm{2}}$
SO, the change in oxidation state$= {\rm{2}}$

$${\rm{Equivalent}}\,{\rm{weight\;of }}{{\rm{O}}_2} = \dfrac{{32\;{\rm{g}}}}{2}\\\ {\rm{Equivalent}}\,{\rm{weight\;of }}{{\rm{O}}_2} = 16\;{\rm{g}}$$

SO, we can write in the form of option as $16\;{\rm{g}}\;{{\rm{O}}_{\rm{2}}} \equiv {\rm{3}}\;{\rm{g}}\;{\rm{C}}$
Now we can calculate of the rest of the options-
(b)${\rm{C + 2}}{{\rm{H}}_{\rm{2}}} \to {\rm{C}}{{\rm{H}}_4}$
Oxidation state of carbon in ${\rm{C}}\,{\rm{ = }}\,{\rm{0}}$
Oxidation state of carbon in ${\rm{C}}{{\rm{H}}_4}{\rm{ = }}\, - {\rm{4}}$
SO, the change in oxidation state ${\rm{ = }}\,{\rm{4}}$
${\rm{equivalent}}\,{\rm{weight\;of\;C}}\,{\rm{ = }}\,\,\dfrac{{1{\rm{2}}\,{\rm{g}}}}{4} = {\rm{3}}\,{\rm{g}}$
Now for hydrogen-
Oxidation state of hydrogen in ${\rm{H}}\,{\rm{ = }}\,{\rm{0}}$
Oxidation state of hydrogen in ${\rm{C}}{{\rm{H}}_4}{\rm{ = }}\, + {\rm{1}}$
SO, the change in oxidation state ${\rm{ = }}\,\,{\rm{1}}$
${\rm{equivalent}}\,{\rm{weight\;of \;H}}\,{\rm{ = }}\,{\rm{1}}\,{\rm{g}}$
SO, we can write in the form of option as ${\rm{1g}}\,{\rm{H}} \equiv {\rm{3g}}\,{\rm{C}}$
Now, for option (c)
(c)${\rm{2C + }}{{\rm{H}}_{\rm{2}}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}$
Oxidation state of carbon in ${\rm{C}}\,{\rm{ = }}\,{\rm{0}}$
Oxidation state of carbon in ${{\rm{C}}_2}{{\rm{H}}_2}{\rm{ = }}\, - {\rm{1}}$
SO, the change in oxidation state ${\rm{ = }}\,{\rm{1}}$
${\rm{equivalent}}\,{\rm{weight\;of\; C}}\,{\rm{ = }}\,\dfrac{{1{\rm{2}}\,{\rm{g}}}}{1} = 12\,{\rm{g}}$
Now for hydrogen-
Oxidation state of hydrogen in ${{\rm{H}}_2}{\rm{ = }}\,{\rm{0}}$
Oxidation state of hydrogen in ${{\rm{C}}_2}{{\rm{H}}_2}{\rm{ = }}\,{\rm{ + 2}}$
SO, the change in oxidation state ${\rm{ = }}\,{\rm{2}}$
${\rm{equivalent}}\,{\rm{weight\;of }}{{\rm{H}}_2}{\rm{ = }}\dfrac{{{\rm{2g}}}}{2}$
SO, we can write in the form of option as ${\rm{1g}}\,{{\rm{H}}_2} \equiv 12\,{\rm{g}}\,{\rm{C}}$
(d)${\rm{2C + 2}}{{\rm{H}}_{\rm{2}}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}$
Oxidation state of carbon in ${\rm{C}}\,{\rm{ = }}\,{\rm{0}}$
Oxidation state of carbon in ${{\rm{C}}_2}{{\rm{H}}_4}{\rm{ = }}\, - {\rm{2}}$
SO, the change in oxidation state ${\rm{ = }}\,{\rm{2}}$
${\rm{equivalent}}\,{\rm{weight\;of\;C}}\,{\rm{ = }}\dfrac{{1{\rm{2}}\,{\rm{g}}}}{2} = 6\,{\rm{g}}$
Now for hydrogen-
Oxidation state of ${{\rm{H}}_2}{\rm{ = }}\,{\rm{0}}$
Oxidation state of ${{\rm{H}}_2}$ in ${{\rm{C}}_2}{{\rm{H}}_4}{\rm{ = }}\,{\rm{ + 2}}$
SO, the change in oxidation state ${\rm{ = }}\,2$
${\rm{equivalent}}\,{\rm{weight\;of }}{{\rm{H}}_2}{\rm{ = }}\dfrac{{2\,g}}{2}$
SO, we can write in the form of option as ${\rm{1g}}\,{{\rm{H}}_2} \equiv 6\,{\rm{g}}\,{\rm{C}}$
Therefore, the options which are given do not match our answer so the correct option is option (1).

Note: We can also use short trick to find our correct option in this type of question as it is very time taking.so, if we find our one answer, just search in other options whether it is given, if given then go to other one answer and if not, then that will be the answer. In Maximum cases, it is applicable.