When 11.2 g of KOH is added to one litre of 'X' M H2SO4 solution, the pOH of the solution becomes 13.301. Then the value of 'X' is (assume no change in the volume of solution by the addition of KOH)
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a
0.4
b
0.6
c
0.8
d
0.2
answer is D.
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Detailed Solution
Following reaction takes placeleft after neutralisation =2×11.256Given, POH = 13.3 ⇒ PH = 0.7⇒ [H+] = 2 ×10-1Thus,⇒ 2x = 4 ×10-1⇒ x = 2 ×10-1