You visited us 1 times! Enjoying our articles? Unlock Full Access!

Raoult's Law

The molar sol...

Question

The molar solubility (in mol $L^{- 1}$ ) of a sparingly soluble salt $M X_{4}$ is $^{'} s^{'}$ . The corresponding solubility product is $K_{s p} . s$ is given in terms of $K_{s p}$ by the relation:

s = (K_{s p} / 128)^{1 / 4}

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

s = (128 K_{s p})^{1 / 4}

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

s = (256 K_{s p})^{1 / 5}

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

s = (K_{s p} / 256)^{1 / 5}

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

Open in App

Solution

The correct option is C $s = (K_{s p} / 256)^{1 / 5}$
$M X_{4} ⇌ M^{4 +} + 4 X^{-}$
$[M X_{4}] = s = [M^{4 +}], [X^{-}] = 4 s$
$K_{s p} = [M^{4 +}] [X^{-}]^{4} = s (4 s)^{4} = 256 s^{5}$
$s = (\frac{K_{s p}}{256})^{\frac{1}{5}}$

Suggest Corrections

Similar questions

The molar solubility ( $m o l L^{- 1})$ of a sparingly soluble salt $M X_{4}$ is $^{'} S^{'}$ . The corresponding solubility product is $K_{S P}$ . $^{'} S^{'}$ is given in terms of $K_{S P}$ by the relation

Q. The molar solubility in

(m o l L^{- 1})

of a sparingly soluble salt

M X_{4}

^{'} s^{'}

.The molar solubility product is

K_{s p}

^{'} s^{'}

is given in terms of

K_{s p}

by the relation:

Q. The molar solubility in

(m o l L^{- 1})

of a sparingly soluble salt

M X_{4}

^{'} s^{'}

.The molar solubility product is

K_{s p}

^{'} s^{'}

is given in terms of

K_{s p}

by the relation:

Q. The molar solubility

(m o l l t^{- 1})

of a sparingly soluble salt

M X_{4}

is `s`. The corresponding solubility product is

K_{s p}

. Then `s`can be given in terms of

k_{s p}

by the relation :

Q. The molar solubility (in

m o l L^{- 1})

of a sparingly soluble salt

M X_{4}

^{'} S^{'}

. The corresponding solubility product is

^{'} K_{s p}^{'}

^{'} S^{'}

in term of

^{'} K_{s p}^{'}

is given by the relation:

Join BYJU'S Learning Program

The molar solubility (in mol L−1) of a sparingly soluble salt MX4 is ′s′. The corresponding solubility product is Ksp. s is given in terms of Ksp by the relation:

The correct option is C s=(Ksp/256)1/5MX4⇌M4++4X−[MX4]=s=[M4+],[X−]=4sKsp=[M4+][X−]4=s(4s)4=256s5s=(Ksp256)15

The molar solubility (in mol $L^{- 1}$ ) of a sparingly soluble salt $M X_{4}$ is $^{'} s^{'}$ . The corresponding solubility product is $K_{s p} . s$ is given in terms of $K_{s p}$ by the relation:

The correct option is C $s = (K_{s p} / 256)^{1 / 5}$
$M X_{4} ⇌ M^{4 +} + 4 X^{-}$
$[M X_{4}] = s = [M^{4 +}], [X^{-}] = 4 s$
$K_{s p} = [M^{4 +}] [X^{-}]^{4} = s (4 s)^{4} = 256 s^{5}$
$s = (\frac{K_{s p}}{256})^{\frac{1}{5}}$