Question

18 gram of Water is mixed in 200 grams of oleum containing 80% free SO3.
What is the new labelling of Oleum?

Answer 1

Answer:

X= 40.

Explanation:

109% is the given % of oleum.

Oleum consists of SO3 and H2SO4

Let SO3 mass in 100g oleum be x …then mass of H2SO4 is (100-x).

Molecular mass of SO3 is 80… , so molesof SO3 are ( x/80 )

The equation of SO3 with water to form H2SO4 is

SO3 +H2O------ > H2SO4.

We can observe that moles of SO3 is same as mole of H2SO4 formed.

So moles of H2SO4 is also x/80 .

Mass of X/80 moles of H2SO4 is 98( x/80 ) . . [ MOLECULAR MASS OF H2SO4 IS 98]

Total mass of H2SO4 in oleum is

98x/80 + ( 100-x )= 109

X= 40.

So percentage of SO3 is (40/100)x 100= 40%

Answer 2

Answer:

Percentage of free SO

3

=20%

20 g SO

3

is present in 100 g oleum.

∵80 g SO

3

requires 18 g H

2

O

20 g SO

3

requires =

80

18

×20=4.5 g H

2

O

∴ Percent labelling of oleum =100+4.5=104.5%

Explanation:

solve like this

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