1.

200 gm of an Oleum sample labeled as 109 % is mixed with 518 gm water then the molality of final mixture is -

Answer»

5.5 m
4 m
4.45 m
1 m

Solution :`(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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