Question 51. The degree of ionisation in water will be ……………. (a) 8 x 10 -7 (b) 0.8 x 10 -9 (c) 3.six x 10 -7 (d) 3.6 x 10 -9 Answer: (a) 8 x 10 -7 Solution: 1 litre of water contains mole. So, degree of ionisation = \(\frac <10^<-7>\times 18><1000>\) = 1.8 x 10 -7
COOH solution?
Question 52. If the solubility product of lead iodide (PbI2) is 3.2 x 10 -8 . Then its solubility in moles/litre will be …………. (a) 2 x 10 -3 . (b) cuatro x 10 -4 (c) 1.6 x 10 -5 (d) 1.8 x 10 -5 Solution: Ksp = 4s 3 4s 3 = 3.2 x 10 -8 s = 2 x 10 -3 M
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