Question

In which of the following electrolytes, largest number of ions are present?

• A. $50g{K}_{2}S{O}_4(mol.wt.=174)$
• B. $50gNaCl(mol.wt.=58.5)$
• C. $50gN{H}_{4}Cl(mol.wt.=53.5)$
• D. $50gCaC{l}_2(mol.wt.=111)$

Answer 1

The correct option is C $50gN{H}_{4}Cl(mol.wt.=53.5)$

The expression for the number of ions is,
Number of ions in the electrolyte $=\frac{\text{Given weight}}{\text{Molecular weight}}\times N_A\times \text{number of ions in one molecule}$

Here, $N_A$ is the Avogadro number.

$K_{2}S{O}_4⟶2K^{+}+{S{O}_4}^{2-}$

It contains $\frac{50\times 3\times N_A}{174}=0.86N_{A}ions$

$NaCl⟶N{a}^{+}+C{l}^{-}$

It contains $\frac{50\times 2\times N_A}{58.5}=1.71N_{A}ions$
$N{H}_{4}Cl⟶{N{H}_4}^{+}+C{l}^{-}$

It contains $\frac{50\times 2\times N_A}{53.5}=1.87N_{A}ions$

$CaC{l}_2⟶C{a}^{2+}+2C{l}^{-}$

It contains $\frac{50\times 3\times N_A}{111}=1.35N_{A}ions$

Hence, option C is correct.