The conductivity of $0.001 M$ $N a_{2} S O_{4}$ solution is $2.6 \times 10^{- 4} {S cm}^{- 1}$ and increases to $7 \times 10^{- 4} {S cm}^{- 1}$ , when the solution is saturated with $C a S O_{4}$ . The molar conductivity of $N a^{+}$ and $C a^{2 +}$ are $50$ and $120 {S cm}^{2} {mol}^{- 1}$ , respectively. Neglect the conductivity of water, find the solubility product of $C a S O_{4}$ .

From the following data answer the questions:
Reaction:

A + B

\to

P

[A] M	[B] M	Initial rate	M sec $^{- 1}$
		$300$ K	$400$ K
$2.5 \times 10^{- 4}$	$3.0 \times 10^{- 5}$	$5.0 \times 10^{- 4}$	$2.0 \times 10^{- 3}$
$5.0 \times 10^{- 4}$	$6.0 \times 10^{- 5}$	$4.0 \times 10^{- 3}$
$1.0 \times 10^{- 3}$	$6.0 \times 10^{- 5}$	$1.6 \times 10^{- 2}$

The value of rate constant at 300 K is

(m^{- 2} s e c^{- 1})

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Enough CaSO4(s) is dissolved in water at 298 K to produce a saturated solution. The concentration of Ca2+ ions is found to be 3.0×10−3M. The Ksp value for CaSO4 will be :

The correct option is A 9.0×10−6[Ca2+]=[SO2−4]=3.0×10−3M Ksp=[Ca2+][SO2−4]=3.0×10−3×3.0×10−3=9.0×10−6

Enough $C a S O_{4} (s)$ is dissolved in water at 298 K to produce a saturated solution. The concentration of $C a^{2 +}$ ions is found to be $3.0 \times 10^{- 3}$ M. The $K_{s p}$ value for $C a S O_{4}$ will be :

The correct option is A $9.0 \times 10^{- 6}$
$[C a^{2 +}] = [S O_{4}^{2 -}] = 3.0 \times 10^{- 3} M$
$K_{s p} = [C a^{2 +}] [S O_{4}^{2 -}] = 3.0 \times 10^{- 3} \times 3.0 \times 10^{- 3} = 9.0 \times 10^{- 6}$