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200 gm of an Oleum sample labeled as 109 % is mixed with 518 gm water then the molality of final mixture is -A. 5.5 mB. 4 mC. 4.45 mD. 1 m |
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Answer» Correct Answer - C `(109) % = (100 + 9) % = 109 gm H_(2)SO_(4)` for each 100 gm Oleum From 200 gm Oleum `H_(2)SO_(4)` produced `= 109 + 109 = 218 gm` Water Remaining `= 518 = 500 gm` Molality `(m) = (218)/(98) xx (1000)/(500) = 4.45 m` |
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