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0.5 g of fuming H2SO4 oleum is diluted with
100 mL of water. This solution is completely
neutralised by 26.7 ml of 0.4 N NaOH. The correct
statement is/are
(1) Mass of so, is 0.104 g
(2) % of free so, = 20.7
(3) Normality of H,SO, for neutralization is 0.2 N
(4) Weight of H,SO, is 0.104 g

Answer 1

Option (c) is the right answer.

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%