Question

Limiting molar conductivities for some ions in water at \(298\,K\) are given below: \[ \lambda_m^\circ(\mathrm{H^+})=349.6,\quad \lambda_m^\circ(\mathrm{Na^+})=50.1,\quad \lambda_m^\circ(\mathrm{Ca^{2+}})=119.0, \] \[ \lambda_m^\circ(\mathrm{Mg^{2+}})=106.0,\quad \lambda_m^\circ(\mathrm{Cl^-})=76.3,\quad \lambda_m^\circ(\mathrm{SO_4^{2-}})=160.0 \] Choose the correct decreasing order of molar conductivity \((\Lambda_m^\circ)\) for \(\mathrm{NaCl}\), \(\mathrm{HCl}\), \(\mathrm{CaCl_2}\) and \(\mathrm{MgSO_4}\).

• A. \(\mathrm{HCl} > \mathrm{CaCl_2} > \mathrm{MgSO_4} > \mathrm{NaCl}\)
• B. \(\mathrm{HCl} > \mathrm{CaCl_2} > \mathrm{NaCl} > \mathrm{MgSO_4}\)
• C. \(\mathrm{NaCl} > \mathrm{CaCl_2} > \mathrm{MgSO_4} > \mathrm{HCl}\)
• D. \(\mathrm{NaCl} > \mathrm{HCl} > \mathrm{CaCl_2} > \mathrm{MgSO_4}\)

Hint:

To find limiting molar conductivity of an electrolyte, simply add the ionic conductivities of all ions produced after dissociation, taking stoichiometric coefficients into account.

Answer 1

The Correct Option is A

Concept: According to Kohlrausch's Law of Independent Migration of Ions, the limiting molar conductivity of an electrolyte is equal to the sum of the limiting ionic conductivities of the ions produced by that electrolyte. \[ \Lambda_m^\circ = \lambda_+^\circ + \lambda_-^\circ \] For electrolytes producing more than one ion of a particular type, the ionic conductivity is multiplied by the corresponding stoichiometric coefficient.

Step 1: Calculate the limiting molar conductivity of HCl.
For HCl, \[ \Lambda_m^\circ(\mathrm{HCl}) = \lambda^\circ(\mathrm{H^+}) + \lambda^\circ(\mathrm{Cl^-}) \] \[ = 349.6+76.3 \] \[ =425.9 \]

Step 2: Calculate the limiting molar conductivity of CaCl\(_2\).
For calcium chloride, \[ \Lambda_m^\circ(\mathrm{CaCl_2}) = \lambda^\circ(\mathrm{Ca^{2+}}) + 2\lambda^\circ(\mathrm{Cl^-}) \] \[ = 119.0+2(76.3) \] \[ = 119.0+152.6 \] \[ =271.6 \]

Step 3: Calculate the limiting molar conductivity of MgSO\(_4\).
\[ \Lambda_m^\circ(\mathrm{MgSO_4}) = 106.0+160.0 \] \[ =266.0 \]

Step 4: Calculate the limiting molar conductivity of NaCl.
\[ \Lambda_m^\circ(\mathrm{NaCl}) = 50.1+76.3 \] \[ =126.4 \]

Step 5: Arrange in decreasing order.
\[ 425.9>271.6>266.0>126.4 \] Hence, \[ \boxed{\mathrm{HCl}>\mathrm{CaCl_2}>\mathrm{MgSO_4}>\mathrm{NaCl}} \]