JEE MAIN - Chemistry (2021 - 18th March Evening Shift - No. 18)
The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm2 mol$$-$$1 respectively. The molar conductivity at infinite dilution of barium sulphate is _________ S cm2 mol$$-$$1. (Round off to the Nearest Integer ).
Answer
288
Explanation
From Kohlrausch's law
$$\Lambda _m^\infty (BaS{O_4}) = \lambda _m^\infty (B{a^{2 + }}) + \lambda _m^\infty (SO_4^{2 - })$$
$$\Lambda _m^\infty (BaS{O_4}) = \Lambda _m^\infty (BaC{l_2}) + \Lambda _m^\infty ({H_2}S{O_4}) - 2\Lambda _m^\infty (HCl)$$
$$ = 280 + 860 - 2(426)$$
$$ = 288$$ S cm2 mol$$-$$1
$$\Lambda _m^\infty (BaS{O_4}) = \lambda _m^\infty (B{a^{2 + }}) + \lambda _m^\infty (SO_4^{2 - })$$
$$\Lambda _m^\infty (BaS{O_4}) = \Lambda _m^\infty (BaC{l_2}) + \Lambda _m^\infty ({H_2}S{O_4}) - 2\Lambda _m^\infty (HCl)$$
$$ = 280 + 860 - 2(426)$$
$$ = 288$$ S cm2 mol$$-$$1
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