NEET-XII-Chemistry
Previous Year Paper year:2018
- #70The solubility of ``BaSO_4`` in water
2.42 × 103 ``gL^{-1}``at 298 K. The value
of solubility product (``K_{sp}``) will be
(Given molar mass of BaSO4 = 233 g ``mol ^{-1}``
(1) ``1.08 × 10^{-10} mol^2 L^{-2}``
(2)`` 1.08 × 10^{-12} mol^2 L^{-2}``
(3)`` 1.08 × 10^{-14} mol^2 L^{-2}``
(4)`` 1.08 × 10^{-8} mol^2 L^{-2}``digAnsr: 1Ans : (1)
Sol. solubility of BaSO4 = 2.42 × 10-3 gL-1
∴ s =
ï€
ï€ ï€ï‚´ = ï‚´
3
5 12.42 10 1.038 10 mol L
233
Ksp = s
2 = (1.038 × 10-5)2
= 1.08 × 10-10 mol2 L-2