25 gram of oleum contains 30 % free so3 . strength of oleum is
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The reaction for the oleum is:
H2SO4 ======> SO3 + H2O
25/98 moles we get 25/98 moles of SO3.
30%of SO3 => 0.3*25/98
Strength => (15/98*2)*100/(25/98) = 50%
H2SO4 ======> SO3 + H2O
25/98 moles we get 25/98 moles of SO3.
30%of SO3 => 0.3*25/98
Strength => (15/98*2)*100/(25/98) = 50%
tvashisht:
No the answer is 106.75%. I only need to know the explanation
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Answer:
given
25gm of oleum
contain 30% free sulphur trioxide (SO3)
strength of oleum ???
The reaction for the oleum is:
H2SO4 -> SO3 + H2O
25/98 moles we get 25/98 moles of SO3.
30%of SO3 => 0.3*25/98
Strength => (15/98*2)*100/(25/98) = 50%
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