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Grade: 11
        
200 gram of an oleum sample (labelled as 109 percent) is mixed with 400 gram of another oleum sample (labelled as 118%).what will be the labelling of new sample formed .
one year ago

Answers : (2)

SAGAR SAHA
13 Points
							
oleum-> So3+H2O->H2SO4                 mole of SO3 = mole of H2O
100 gm oleum + 9 gm H2O                    wt. of SO3/80 =18/18
200 gm oleum + 18gm H2O                   wt. of SO3 = 80 gm
 
400 gm oleum ->118%
100 g → 18 H2O
400 g → 4*18 H2O → 72g H2O
wt of SO3/80 = 72/18
wt of SO3= 320g 
H2SO4= 400-320= 80g
 Total wt of SO3= 320+80=400g
 
SO3+H2O → H2SO4
MOLE OF SO3 = MOLE OF H2O 
400/80= wt OF H2O/18
wt OF H2O= 400*18/80 = 90gm 
600 gm OLEUM → 90 gm H2O
100 gm OLEUM → 90/6 = 15 gm H20 
 
100+ wt of H2O 
= 100+15
= 115% (ANSWER)
2 months ago
Arun
24735 Points
							
Dear student
 
 
oleum-> So3+H2O->H2SO4                 mole of SO3 = mole of H2O
100 gm oleum + 9 gm H2O                    wt. of SO3/80 =18/18
200 gm oleum + 18gm H2O                   wt. of SO3 = 80 gm
 
400 gm oleum ->118%
100 g → 18 H2O
400 g → 4*18 H2O → 72g H2O
wt of SO3/80 = 72/18
wt of SO3= 320g 
H2SO4= 400-320= 80g
 Total wt of SO3= 320+80=400g
 
SO3+H2O → H2SO4
MOLE OF SO3 = MOLE OF H2O 
400/80= wt OF H2O/18
wt OF H2O= 400*18/80 = 90gm 
600 gm OLEUM → 90 gm H2O
100 gm OLEUM → 90/6 = 15 gm H20 
 
100+ wt of H2O 
= 100+15= 115 percent
 
Hope it helps
 
2 months ago
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