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

Options

A. 5.5 m
B. 4 m
C. 4.45 m
D. 1 m

Answer

C

Explanation

`(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`