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9g water is added into 100g of oleum sample labelled as 112% H2SO4, then the amount of free SO3 remaining in the solution is? (Working pls)

9g water is added into 100g of oleum sample labelled as 112% H2SO4, then the amount of free SO3 remaining in the solution is? (Working pls)

Grade:11

1 Answers

Vikas TU
14149 Points
4 years ago
The condition for response is : 
 
SO3+H2O— — −>H2SO4 
 
This infers the measure of moles of water responds with same number of moles of SO3 in the specimen : 
 
The quantity of moles of water included is = MASS/MOLECULAR MASS =9/18= 0.5 mole of water 
 
This responds with an equivalent number of moles of SO3 : 
 
Unique number of moles of SO3 in the specimen is = mass/atomic mass 
 
Mass is computed as: 
 
Be that as it may, the quantity of moles of H2S04 framed would be : y/80 
 
The aggregate mass of H2SO4 would be mole X atomic mass = 98y/80 
 
Mass of H2 SO4 in 100 gm would be 100-y 
 
Be that as it may, new mass framed + existing mass must be equivalent to 109.9% henceforth ; 
 
98y/80 + (100-y) = 100.9 
 
Y= 53.33 gm 
 
Number of moles is 53.33/80 
 
= 0.6667 
 
Number of moles staying after response with water : 
 
= 0.6667-0.5 
 
=0.1667 
 
The mass that remaining parts is = moles X sub-atomic mass 
 
= 0.1667 X 80 
 
= 13.33 gm

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