Verified Solution • 2025–26 Session moderate

QJEEM19Ch456b: JEE Main 2019 Chemistry - Class 12 JEEM19Ch456b Solution: for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm 2 mol –1…

∧m∘\wedge _m^ \circ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is-

5 ×\times 10–5 S cm–1, degree of dissociation of HA is -

Multiple Choice Options

  • A 0.5
  • B 0.125
  • C 0.25
  • D 0.75

Step-by-Step Verified Solution

∧m∘\wedge _m^ \circ (HA) = ∧m∘\wedge _m^ \circ (HCl) + ∧m∘\wedge _m^ \circ (NaA) −- ∧m∘\wedge _m^ \circ (NaCl)

= 425.9 + 100.5 −- 126.4

= 400 S cm2 . mol −- 1

     ∧mc\wedge _m^ c = K×1000M{{K \times 1000} \over M}

= 5×10−5×100010−3{{5 \times {{10}^{ - 5}} \times 1000} \over {{{10}^{ - 3}}}}

= 50 S cm2 mol −- 1

α\alpha = ΛmcΛmo{{\Lambda _m^c} \over {\Lambda _m^o}} = 50400{{50} \over {400}} = 0.125

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