JEE MAIN - Chemistry (2022 - 25th June Evening Shift - No. 4)
The Ksp for bismuth sulphide (Bi2S3) is 1.08 $$\times$$ 10$$-$$73. The solubility of Bi2S3 in mol L$$-$$1 at 298 K is :
1.0 $$\times$$ 10$$-$$15
2.7 $$\times$$ 10$$-$$12
3.2 $$\times$$ 10$$-$$10
4.2 $$\times$$ 10$$-$$8
Explanation
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Ksp = (2s)2(3s)3 = 108s5
108s5 = 108 × 10–75
s = 1.0 × 10–15 mol/L
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