Question

Calculate molar conductivity of $HCOOH$ at infinite dilution, if equivalent conductivity of $H_{2}S{O}_4=x_1,A{l}_2(S{O}_4{)}_2=x_2,(HCOO{)}_{3}Al=x_3$.

• A. $6x_1-3x_2+6x_3$
• B. $\frac{x_1-x_2+x_3}{6}$
• C. $x_1-x_2+x_3$
• D. $\frac{6x_1-3x_2+6x_3}{6}$

Answer 1

The correct option is C $x_1-x_2+x_3$

${\wedge }_{eq}=\left(\frac{1}{n^{+}\times z^{+}}\right){\wedge }_m$ ${\wedge }_{eq}$= Equivalent conductivity; ${\wedge }_m$= molar conductivity; $n^{+}$= no. of cation; $z^{+}$= charge of cation

$H+2S{O}_4⟶2H^{+}+S{O}_4^{2-}$, ${\wedge }_{eq}^{\prime }=x_1⟶\left(1\right)$

$\Rightarrow {\wedge }_{eq}^{\prime }=\frac{1}{(2\times 1)}{\wedge }^{\prime }m\Rightarrow {\wedge }^{\prime }m=2x_1$

$A{l}_2(S{O}_4{)}_3⟶3A{l}^{3+}+3S{O}_4^{2-}$ ${\wedge }_{eq}^{\prime \prime }=x_2⟶\left(2\right)$

$\Rightarrow \wedge {"}_{eq}=\frac{1}{(2\times 3)}{\wedge }^{\prime \prime }m$

$\Rightarrow {\wedge }^{\prime \prime }m=6x_2$

$(HCOO{)}_{3}Al⟶A{l}^{3+}+3HCO{O}^{-}$ ${\wedge }_{eq}^{\prime \prime \prime }=x_3⟶\left(3\right)$

$\Rightarrow {\wedge }_{eq}^{\prime \prime \prime }=\frac{1}{(1\times 3)}{\wedge }^{\prime \prime \prime }m\Rightarrow {\wedge }^{\prime \prime \prime }m=3x_3$

On multiply the equation $\left(1\right)$ by $3$ & equation $\left(3\right)$ by $2$ we get,

$3H_{2}S{O}_4⟶6H^{+}+3S{O}_4^{2-}$ ${\wedge }_{eq}^{\prime }=x_1⟶\left(4\right)$

${\wedge }^{\prime }m=6x_1$

$2(HCOO{)}_{3}Al⟶2A{l}^{3+}+6HCO{O}^{-}$ ${\wedge }_{eq}^{\prime \prime \prime }=x_3$

${\wedge }^{\prime \prime \prime }m=(2\times 3)x_3=6x_3⟶\left(5\right)$

On adding equation $4$ and $5$ subtracting $2$ from them we get

$A{l}_2(S{O}_4{)}_3+3H_{2}S{O}_4+2(HCOO{)}_{3}Al⟶6H^{+}+6HCO{O}^{-}$

${\wedge }^{\prime \prime }m={\wedge }^{\prime }m{\wedge }^{\prime \prime }m+{\wedge }^{\prime \prime \prime }m$

$\wedge m=6x_1-6x_2+6x_3⟶\left(6\right)$

For $HCOOH\rightleftharpoons H^{+}+HCO{O}^{-}$

$\wedge m=(1\times 1){\wedge }_{eq}$

Thus the equation $6$ will be divided by $6$ as $6H^{+}$ ions are produced.

$\wedge m=x_1-x_2+x_3$