The molality of 49% by volume of `H_(2)SO_(4)` solution having density 1.49 g/mL is……

Answer

5

Explanation

49% by volume means
49 g of `H_(2)SO_(4)` is present in 100 mL solution
Mass of solution = Volume `xx d = 100 xx 1.49 g = 149 g`
Mass of solvent = Mass of solution - mass of solute = 149 - 49 = 100g
Molality `= (n_(B))/(g_(A)) xx 100 , " "n_(B) = (g_(B))/(M_(0)(B)) = (49)/(98) 0.5 " ":. "Molality" = (0.5)/(100) xx 1000 = 5`