Question

The equivalent conductance of $B{a^{2 + }}$ and $C{l^ - }$ are respectively $127\& 76oh{m^{ - 1}}c{m^2}e{q^{ - 1}}$ at infinite dilution. What will be the equivalent conductance of $BaC{l_2}$ at infinite dilution?
A. $139.5oh{m^{ - 1}}c{m^2}e{q^{ - 1}}$
B. $203oh{m^{ - 1}}c{m^2}e{q^{ - 1}}$
C. $279oh{m^{ - 1}}c{m^2}e{q^{ - 1}}$
D. $101.5oh{m^{ - 1}}c{m^2}e{q^{ - 1}}$

Answer 1

We know that the conducting power of all the ions produced by one gram equivalent of electrolyte is termed as equivalent conductance.
According to Kohlrausch’s law, at infinite dilution the sum of the equivalence conductance of ions present in the component is equal to the equivalent conductance of electrolyte

Complete step by step answer: We can write the chemical reaction as,
$B{a^{2 + }} + 2C{l^ - }\xrightarrow{{}}BaC{l_2}$
We define the equivalent weight of the substance as the ratio of the molecular weight or mass of the compound to the n-factor or the acidity or basicity. We can calculate the n- factor by determining the change in oxidation state.
The equivalent weight of $BaC{l_2} = \dfrac{{{\text{Molecular weight}}}}{2}$
The equivalent conductance of barium chloride can be calculated as,
$\lambda _m^\infty BaC{l_2} = \lambda _m^\infty \left( {B{a^{2 + }}} \right) + 2\lambda _m^\infty \left( {C{l^ - }} \right)$
Substituting the known values we get,
$\Rightarrow \lambda _m^\infty BaC{l_2} = \dfrac{1}{2} \times 127 + \left( {2 \times 76} \right)$
$\lambda _m^\infty BaC{l_2} = \dfrac{{127}}{2} + 152$
On simplification we get,
$\Rightarrow \lambda _m^\infty BaC{l_2} = 139.5oh{m^{ - 1}}c{m^2}e{q^{ - 1}}$
Example:
The equivalent weight of the sulfur dioxide is calculated as,
We know the molecular mass of the sulfur dioxide is $64g/mol$.
Now, we calculate the n-factor. First, calculate the change in the oxidation state of sulfur in the reactions.
The given reactions is,
$S{O_2} + 2{H_2}S = 3S + 2{H_2}O$
The oxidation state of sulfur is changed from $+ 4$ to zero. Thus the n-factor is $4.$
The equivalent weight of sulfur$= \dfrac{{64}}{4} = 16g$
$5S{O_2} + 2KMn{O_4} + 2{H_2}O = {K_2}S{O_4} + 2MnS{O_4} + 2{H_2}S{O_4}$
The oxidation state of sulfur is changed from $+ 4$ to$+ 6$. Thus the n-factor is$2$.
The equivalent weight of sulfur$= \dfrac{{64}}{2} = 32g$
Thus, the equivalent weight of sulfur in reaction a is $16g$ and the equivalent weight of sulfur in reaction a is $32g$.
So, the correct answer is “Option A”.

Note:
We can also define the equivalent weight of the substance as the ratio of the molecular weight or mass of the compound to the n-factor or the acidity or basicity. We can calculate the n- factor by determining the change in oxidation state. The 'n' factor of an acid is the number of ions replaced by one mole of acid. The n-factor for acid isn't the number of moles of usable Hydrogen atoms present in one mole of acid. The oxidation state of the single element is zero but if it has any charge present on it then it is considered as n-factor while basicity for the acidic substance and the acidity is defined for the basic substance.