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[" 0.5g of fuming "H(2)SO(4)" (oleum) is...

[" 0.5g of fuming "H_(2)SO_(4)" (oleum) is diluted with water.This solution is completely "],[" neutralized by "26.7ml" of "0.4N" NaOH.the "%" of free "SO_(3)" in the sample "],[" is "]

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0.5 gm of fuming H_(2)SO_(4) (Oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 M NaOH solution. Calculate the percentage of free SO_(3) in the given sample. Give your answer excluding the decimal places.

0.5 gm of fuming H_(2)SO_(4) (Oleum) is diluted with water. This solution is completely neutralised by 26.7 ml of 0.4 M NaOH solution. Calculate the percentage of free SO_(3) in the given sample. Give your answer excluding the decimal places.

Oleum or fuming H_(2)SO_(4) is

0.5 g of fuming sulphuric acid (H_2SO_4+SO_3) , called oleum, is diluted with water. Thus solution completely neutralised 26.7 " mL of " 0.4 M NaOH . Find the percentage of free SO_3 in the sample solution.

0.5 g of fuming sulphuric acid (H_2SO_4+SO_3) , called oleum, is diluted with water. Thus solution completely neutralised 26.7 " mL of " 0.4 M NaOH . Find the percentage of free SO_3 in the sample solution.