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 .

Question

SAGAR SAHA · Answer

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)

Arun · Answer

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

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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 .

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 .

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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 .

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 .

2 Answers

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