Q 1210423319.     Calculate Its solubility product at `298 K, `

At `298 K`, the conductivity of a saturated solution of `AgCl` in water is `2.6 × 10^(-6) S cm^(-1)`.

( given :` λ^(∞) (Ag^+ ) = 63.0 S cm^2mol^(-1) , λ^(∞) (Cl^(∞)) = 67.0 S cm^(2)Mol^(-1) `)

A

`2.0 × 10^(-5) M^2`

B

`2 × 10^(-8) M^2`

C

`4.0 × 10^(-16 ) M^2`

D

`4.0 × 10^(-10) M^2`

HINT

Use formula: Solubility `S =(1000k)/(λ_(AgCl)^(0))`

Find instant solution to any question from our database of 1.5 Lakh + Questions

 

Access free resources including

Engineering

Medical

Banking

DEFENCE

SSC

RRB

CBSE-NCERT

OLYMPIAD

Today's Score

0
Success Rate: NAN %

Practice & Test

Sr. No.SubjectTagCorrectRecorded Answer