The values of Lambdae q∞ for NH4Cl , NaOH and NaCl are, respectively, 149.74, 248.1 and 126.4 ohm- 1cm2eq- 1 . The value of Lambdae q∈ ftyNH4OH is (. in ohm . -1 cm 2 eq -1)

Question

The values of Lambdae q∞ for NH4Cl , NaOH and NaCl are, respectively, 149.74, 248.1 and 126.4 ohm- 1cm2eq- 1 . The value of Lambdae q∈ ftyNH4OH is (. in ohm . -1 cm 2 eq -1) (A) 371.44 (B) 271.44 (C) 71.44 (D) It cannot be calculated from the data given

Tardigrade Admin · Accepted Answer

Lambda text eq ∞ for NH 4 Cl can be find out by the relation given below Lambda eq 0 or Lambda∞( NH 4 OH )= Lambda∞ eq ( NH 4 Cl )+ Lambda∞ eq ( NaOH )- Lambda∞ eq ( NaCl ) =(149.74+248.1-126.4) =271.44 ohm -1 cm 2 eq -1

Q. The values of $Λ_{e q}^{\infty}$ for $N H_{4} Cl$ , NaOH and NaCl are, respectively, 149.74, 248.1 and 126.4 $o h m^{- 1} c m^{2} e q^{- 1}$ . The value of $Λ_{e q}^{\in f t y} N H_{4} O H$ is
$($ in ohm $^{- 1} c m^{2} e q^{- 1})$

371.44

271.44

71.44

It cannot be calculated from the data given

$Λ_{eq}^{\infty}$ for $N H_{4} Cl$ can be find out by the relation given below
$Λ_{e q}^{0}$ or $Λ^{\infty} (N H_{4} O H) = Λ^{\infty} e q (N H_{4} Cl) + Λ^{\infty} e q (N a O H) - Λ^{\infty} e q (N a Cl)$
$= (149.74 + 248.1 - 126.4)$
$= 271.44 o h m^{- 1} c m^{2} e q^{- 1}$

Q. The values of Λeq∞​ for NH4​Cl , NaOH and NaCl are, respectively, 149.74, 248.1 and 126.4 ohm−1cm2eq−1 . The value of Λeq∈fty​NH4​OH is ( in ohm −1cm2eq−1)