Given the following molar conductivites at 25- C - HCl -426 -1 cm 2 mol -1 - NaCl -126 -1 cm 2 mol -1- NaCro sodium crotonate -83 -1 cm 2 mol -1- The conductivity of 0-001 mol - dm 3 acid solution is 3-83 - 10-5-1 cm -1- The dissociation constant of crotonic acid isA- 4-11 - 10-5 MB- 3-11 - 10-5 MC- 2-11 - 10-5 MD- 1-11 - 10-5 M

The correct option is D 1.11 × 10^ 5M Λ^∘=426 126+83=383 Λ=3.83× 10^ 5×1000/0.001=38.3 ∝ =Λ/Λ^∘=38.3/383 = 0.1; K_a=C∝^2/1 ∝=10^ 3× (0.1)^2/(1 0.1)=1.11× 10^ 5 ...

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Standard X

Physics

Resistance

Given the fol...

Question

Given the following molar conductivites at $25^{\circ} C$ ; $H C l = 426 Ω^{- 1} c m^{2} m o l^{- 1}$ ; $N a C l = 126 Ω^{- 1} c m^{2} m o l^{- 1}$ , NaCro (sodium crotonate) $= 83 Ω^{- 1} c m^{2} m o l^{- 1}$ . The conductivity of $0.001 m o l / d m^{3}$ acid solution is $3.83 \times 10^{- 5} Ω^{- 1} c m^{- 1}$ . The dissociation constant of crotonic acid is

$4.11 \times 10^{- 5} M$

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$3.11 \times 10^{- 5} M$

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$2.11 \times 10^{- 5} M$

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$1.11 \times 10^{- 5} M$

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Solution

The correct option is D
$1.11 \times 10^{- 5} M$

$Λ^{\circ} = 426 - 126 + 83 = 383$

$Λ = 3.83 \times 10^{- 5} \times \frac{1000}{0.001} = 38.3$

$\propto= \frac{Λ}{Λ^{\circ}} = \frac{38.3}{383} = 0.1; K_{a} = \frac{C \propto^{2}}{1 - \propto} = \frac{10^{- 3} \times (0.1)^{2}}{(1 - 0.1)} = 1.11 \times 10^{- 5}$

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Similar questions

Q. Given the following molar conductivities at

25^{\circ} C; H C l, 426 Ω^{- 1} c m^{2} m o l^{- 1}; N a C l,

126 Ω^{- 1} c m^{2} m o l^{- 1}; N a C (s o d i u m c r o t o n a t e)

83 Ω^{- 1} c m^{2} m o l^{- 1}

. What is the ionization constant of crotonic acid? If the conductivity of a

0.001 M

crotonic acid solution is

3.83 \times 10^{- 5} Ω^{- 1} c m^{- 1}

Q. What is the ionization constant of crotonic acid if the conductivity of a

0.001 M

crotonic acid solution is

3.83 \times 10^{- 5} Ω^{- 1} c m^{- 1}

?

Limiting molar conductivities at

25^{o} C

H C l = 426 Ω^{- 1} c m^{2} m o l^{- 1}

N a C l = 126 Ω^{- 1} c m^{2} m o l^{- 1}

N a C (sodium crotonate) = 83 Ω^{- 1} c m^{2} m o l^{- 1}

.
(Crotonate ion is represented by

C^{-}

)

Q. Given the limiting molar conductivity as

_{m}

(HCl)

= 425.9 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

(NaCl)

= 126.4 Ω^{- 1} c m^{2} m o l^{- 1}

_{m}

(CH3 COONa)

= 91 Ω^{- 1} c m^{2} m o l^{- 1}

The molar conductivity, at infinite dilution, of acetic acid (in

Ω^{- 1} c m^{2} m o l^{- 1}

) will be

Q. The conductivity of

0.01 m o l / d m^{3}

aqueous acetic acid at

300 K

19.5 \times 10^{- 5} o h m^{- 1} c m^{- 1}

and the limiting molar conductivity of acetic acid at the same temperature is

390 o h m^{- 1} c m^{2} m o l^{- 1}

. The degree of dissociation of acetic acid is:

Q. The specific conductance of a

0.20 m o l L^{- 1}

solution of an electrolyte at

20^{0} C i s 2.48 \times 10^{- 4} o h m^{- 1} c m^{- 1}

. The molar conductivity of the solution is :

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Given the following molar conductivites at 25∘ C; HCl=426Ω−1 cm2 mol−1; NaCl=126Ω−1 cm2 mol−1, NaCro (sodium crotonate) =83Ω−1 cm2 mol−1. The conductivity of 0.001 mol/dm3 acid solution is 3.83×10−5Ω−1 cm−1. The dissociation constant of crotonic acid is

The correct option is D 1.11×10−5M Λ∘=426−126+83=383 Λ=3.83×10−5×10000.001=38.3 ∝=ΛΛ∘=38.3383=0.1;Ka=C∝21−∝=10−3×(0.1)2(1−0.1)=1.11×10−5