JEE MAIN - Chemistry (2019 - 12th January Evening Slot - No. 23)
$$ \wedge _m^ \circ $$ 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 -
5 $$ \times $$ 10–5 S cm–1, degree of dissociation of HA is -
0.50
0.125
0.25
0.75
Explanation
$$ \wedge _m^ \circ $$ (HA) = $$ \wedge _m^ \circ $$ (HCl) + $$ \wedge _m^ \circ $$ (NaA) $$-$$ $$ \wedge _m^ \circ $$ (NaCl)
= 425.9 + 100.5 $$-$$ 126.4
= 400 S cm2 . mol$$-$$1
$$ \wedge _m^ c $$ = $${{K \times 1000} \over M}$$
= $${{5 \times {{10}^{ - 5}} \times 1000} \over {{{10}^{ - 3}}}}$$
= 50 S cm2 mol$$-$$1
$$\alpha $$ = $${{\Lambda _m^c} \over {\Lambda _m^o}}$$ = $${{50} \over {400}}$$ = 0.125
= 425.9 + 100.5 $$-$$ 126.4
= 400 S cm2 . mol$$-$$1
$$ \wedge _m^ c $$ = $${{K \times 1000} \over M}$$
= $${{5 \times {{10}^{ - 5}} \times 1000} \over {{{10}^{ - 3}}}}$$
= 50 S cm2 mol$$-$$1
$$\alpha $$ = $${{\Lambda _m^c} \over {\Lambda _m^o}}$$ = $${{50} \over {400}}$$ = 0.125
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