0 g of fuming h2so4 is diluted with water. This solution is completely neutralised by 26 ml of 0.8 n naoh. The % of free so3 in the sample is
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0.5g of fuming H2SO4 is diluted with water. The solution requires 26.7ml of 0.4 N NAOH for complete neutralization. Can you find the percentage of free SO3?
Let number of eq.of SO3 = x/40, number of eq.of H2SO4 = (0.5-x)/49 so,
(x/40) + (0.5-x)/49 = 0.4×26.7/1000, on solving we get x value.
% of free sulphur tri-oxide in the sample = (100x/0.5)
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H2So4= 98/2 = 49
So3 = 80/2 = 40
26.7ml x 0.4N = 10.68ml - 0.01068L
x/49 + x/40 (0.5-X)
Eq. H2SO4 = Eq of NaOH
= (26.7 x 0.4)/1000
= 10.68 mEq.
H2SO4 + SO3 + H2O -----> 2(H2SO4)
==> SO3 + H2SO4 -----> H2SO4
∴Eq of SO3 = 1/2 of mEq. total H2SO4
==> Eq of H2SO4 + Eq of SO3 = Eq of total H2SO4
= 10.68/1000 (bec. 1 mEq. = 1/1000 Eq.)
Let x be mass of SO2 ==> mass of H2SO4
= (0.5-x)
= (0.5 - x)/49 + x/40 = 10.68/1000
Solving.... x = 0.1836g
Percentage of So3 = 0.1836/0.5 x 100
= 20.73%
Hope this helps you mate
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