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