Arrange `H_(2)CO_(3)` (carbonic acid), glucose, sodium chloride and hydrochloric acid solution, each of 0.1 (M) concentration, in ascending order of their molar conductivity.

Explanation

At a deninite temperature the molar conductivity `(Lambda_(m))` of an electrolyte solution is proportional to - (i) number of ions in the solution (ii) valency of the ions (iii) ionic mobility.
Among the given electrolytes, glucose is a nonelectrolyte. Hence, its solution is a weak electrolyte, whereas `NaCl` and `HCl,` carbonic acid is a weak electrolyte, whereas `NaCl and HCl` are strong electrolytes. So, number of ions in 0.1 M `H_(2)CO_(3)` solution will be fewer than that is 0.1 M NaCl or 0.1 M HCl solution.
As `NaCl` and `HCl` are strong electrolytes, they dissociate almost completely in their solutions. So, the solutions of NaCl and KCl with equal concentration will contains the same number of ions. Both solution contains `Cl^(-)` ions, therefore, their differences in molar conductivities arise due to difference in ionic conductance of `Na^(+)` and `H^(+)` ions. As `H^(+)` ion has higher ionic conductivity than `Na^(+)` ion, the molar conductivity 0f 0.1 M of HCl solution will be greater than that of 0.1 M `NaCl` solution. Therefore, arranging the given solutions in order of their increasing molar conductivity will give the sequence:
`Lambda_(m)"(glucose)" lt Lambda_(m)(H_(2)CO_(3))lt Lambda_(m)(NaCl)lt Lambda_(m)(HCl)`