Conductivity of a saturated solution of a sparingly soluble salt AB at 298 K is 1-85- 10 -5 S m -1- Solubility product of the salt AB at 298 K is -Given - m o AB -140- 10 -4 S m 2 mol -1-

The correct option is B 1.74× 10 ^ 12 Λ_m^0=k×1000C 140×10^ 4=1.85×10^ 5×11000C as units are in metres C≃1.32×10^ 6 AB ⇋ A^+ + B^ ...

217

You visited us 217 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Conductance of Electrolytic Solution

Conductivity ...

Question

Conductivity of a saturated solution of a sparingly soluble salt $A B$ at $298 K$ is $1.85 \times 10^{- 5} S m^{- 1}$ . Solubility product of the salt $A B$ at $298 K$ is :

[Given $Λ_{m}^{o} (A B) = 140 \times 10^{- 4} S m^{2} {m o l}^{- 1}$ ]

1.32 \times 10^{- 12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.74 \times 10^{- 12}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

5.7 \times 10^{- 12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.5 \times 10^{- 12}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $1.74 \times 10^{- 12}$
$Λ_{m}^{0} = k \times \frac{1000}{C}$

$140 \times 10^{- 4} = 1.85 \times 10^{- 5} \times \frac{1}{1000 C}$ as units are in metres

$C ≃ 1.32 \times 10^{- 6}$

$A B ⇋ A^{+} + B^{-}$

$K_{s p} = C^{2} = 1.74 \times 10^{- 12}$

Hence, option B is correct.

Suggest Corrections

Similar questions

Q. Specific conductance of saturated solution of

A g C l

2.68 \times 10^{- 4} S m^{- 1}

and that of water is

0.86 \times 10^{- 4} S m^{- 1}

298 K

. Calculate the solubility product in

m o l / {d m}^{3}

unit if

λ_{m}^{o}

for

H {N O}_{3}, A g {N O}_{3}

and

H C l

are

421 \times 10^{- 4} S m^{2}

{m o l}^{- 1}

144 \times 10^{- 4} S m^{2}

{m o l}^{- 1}

426 \times 10^{- 4} S m^{2}

{m o l}^{- 1}

respectively:

Q. Solubility product

K_{s p}

of a sparingly soluble salt

Z n (C N)_{2}

8 \times 10^{- 12} .

Its solubility value in

(m o l L^{- 1})

is:
(Take

\sqrt[3]{2} = 1.26

)

Q. At

298 K

, the conductivity of a saturated solution of

A g C l

in water is

2.6 \times 10^{- 6} S {c m}^{- 1}

. Its solubility product at

298 K

is:

[Given:

λ^{\infty} ({A g}^{+}) = 63.0 S {c m}^{2} {m o l}^{- 1}

λ^{\infty} ({C l}^{-}) = 67.0 S {c m}^{2} {m o l}^{- 1}

]

Q. The electrochemical cell shown below is a concentration cell,

M | M^{2 +}

(Saturated solution of a sparingly soluble salt,

M X_{2}) | | M^{2 +} (0.001 m o l d m^{- 3}) M .

The emf of the cell depends on the difference in concentration of

M^{2 +}

ions at the two electrodes. The emf of the cell at 298K is 0.059

The solubility product

(K_{s p} : m o l^{3} d m^{- 9})

M X_{2}

at 298 K based on the information available the given concentration cell is (take

2.303 \times R \times 298 / F = 0.059 V)

Q. Solubility product

K_{s p}

of a sparingly soluble salt

Z n (C N)_{2}

8 \times 10^{- 12} .

Its solubility value in

(m o l L^{- 1})

is:
(Take

\sqrt[3]{2} = 1.26

)

Join BYJU'S Learning Program

Conductivity of a saturated solution of a sparingly soluble salt AB at 298K is 1.85×10−5Sm−1. Solubility product of the salt AB at 298K is :[Given Λom(AB)=140×10−4Sm2mol−1]

The correct option is B 1.74×10−12Λ0m=k×1000C140×10−4=1.85×10−5×11000C as units are in metresC≃1.32×10−6AB⇋A++B−Ksp=C2=1.74×10−12Hence, option B is correct.

Conductivity of a saturated solution of a sparingly soluble salt $A B$ at $298 K$ is $1.85 \times 10^{- 5} S m^{- 1}$ . Solubility product of the salt $A B$ at $298 K$ is :

[Given $Λ_{m}^{o} (A B) = 140 \times 10^{- 4} S m^{2} {m o l}^{- 1}$ ]