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Standard XII

Chemistry

Equivalent, Molar Conductivity and Cell Constant

Calculate Δ...

Question

Calculate $Δ_{H O A c}^{\infty}$ $(i n S c m^{2} m o l^{- 1})$ using appropriate molar conductances of the electrolytes listed in the given table at infinite dilution in $H_{2} O$ at $25^{0}$ C.

Electrolyte KCl KNO $_{3}$ HCl NaOAc NaCl
$Δ^{\infty} (S c m^{2} m o l^{- 1}$ 149.9 145.0 426.2 91.0 126.5

517.2

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305

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390.7

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217.5

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Solution

The correct option is D 390.7
Reaction:
$H C l + N a O A c ⟶ N a C l + H O A c$

Now,
$Λ_{H O A c}^{\infty} = Λ_{N a O A c}^{\infty} + Λ_{H C l}^{\infty} - Λ_{N a C l}^{\infty}$

$= (91.0 + 426.2 - 126.5) S c m^{2} m o l^{- 1}$

$= 390.7 S c m^{2} m o l^{- 1}$

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Similar questions

Electrolyte $\land^{\infty} (S c m^{2} m o l^{- 1})$

KCl 149.9

KNO $_{3}$ 145.0

HCl 426.2

NaOAc 91.0

NaCl 126.5

Calculate $\land {^{\infty}}_{H O A c}$ using appropriate molar conductances of the electrolytes listed above at infinite dilution in $H_{2} O$ at 25 $^{0}$ C:

Electrolyte	$K C l$	$K {N O}_{3}$	$H C l$	$N a O A c$	$N a C l$
$Λ^{\infty} (S {c m}^{2} {m o l}^{- 1})$	$149.9$	$145.0$	$426.2$	$91.0$	$126.5$

Calculate

{Λ^{\infty}}_{H O A c}

using appropriate molar conductances of the electrolytes listed above at infinite dilution in

H_{2} O

25^{o} C

Electrolyte $\land^{\infty} (S c m^{2} m o l^{- 1})$

$K C l$ 149.9

$K N O_{3}$ 145.0

$H C l$ 426.2

$N a O A c$ 91.0

$N a C l$ 126.5

Calculate $Λ {^{\infty}}_{H O A c}$ using appropriate molar conductances of the electrolytes listed above at infinite dilution with $H_{2} O$ at 25 $^{o}$ C?

Q. Calculate

Λ_{H N O_{3}}^{0}

and

Λ_{H O A c}^{0}

using appropriate molar conductances of the electrolytes listed below at infinite dilution in

H_{2} O

25^{\circ} C

Electrolyte	$Λ^{0} (S c m^{2} m o l^{- 1})$
$K C l$	$149.9$
$N a C l$	$126.5$
$H C l$	$426.2$
$K N O_{3}$	$145.0$
$N a O A c$	$91.0$

Electrolyte	$K C l$	$K N O_{3}$	$H C l$	$N a O A c$	$N a C l$
$Λ^{\infty} {(Scm}^{2} {mol}^{- 1})$	$149.9$	$145$	$426.2$	$91$	$126.5$

Calculate

Λ_{H O A c}^{\infty}

using appropirate molar conductances of the electrolytes listed above at infinite dilution in

H_{2} O

25^{o} C

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Electrolyte	KCl	KNO $_{3}$	HCl	NaOAc	NaCl
$Δ^{\infty} (S c m^{2} m o l^{- 1}$	149.9	145.0	426.2	91.0	126.5

Calculate Δ∞HOAc (inScm2mol−1) using appropriate molar conductances of the electrolytes listed in the given table at infinite dilution in H2O at 250 C.ElectrolyteKClKNO3HClNaOAcNaClΔ∞(Scm2mol−1149.9145.0426.291.0126.5

The correct option is D 390.7Reaction:HCl+NaOAc⟶NaCl+HOAcNow,Λ∞HOAc=Λ∞NaOAc+Λ∞HCl−Λ∞NaCl =(91.0+426.2−126.5) Scm2mol−1 =390.7 Scm2mol−1

The correct option is D 390.7
Reaction:
$H C l + N a O A c ⟶ N a C l + H O A c$

Now,
$Λ_{H O A c}^{\infty} = Λ_{N a O A c}^{\infty} + Λ_{H C l}^{\infty} - Λ_{N a C l}^{\infty}$

$= (91.0 + 426.2 - 126.5) S c m^{2} m o l^{- 1}$

$= 390.7 S c m^{2} m o l^{- 1}$