Class 11 NEET Chemistry Equilibrium Questions Part 1
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The solubility of B a S O 4 in water is 2.42 × 10 − 3 g L − 1 at 298 K. The value of its solubility product ( K s p ) will be
(Given molar mass ofB a S O 4 = 233 g m o l − 1 )
(Given molar mass of
[NEET 2018]
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