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0.5 g of fuming `H_2SO_4` (oleum) is diluted with water. This solution is completely neutralized by 26.7 ml of 0.4 N NaOH. The percentage of free `SO_3` in the sample is
A. |
30.6% |
B. |
40.6% |
C. |
20.6% |
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D. |
50% |
% of `SO_3=(0.103)/0.5xx100=20.6%`