What volume of water is required (in mL) to prepare 1 L of 1 M solution of `H_(2)SO_(4)` (density = 1.5g/mL) by using 109% oleum and water only (Take density of pure water = 1g/mL).

Explanation

1 mole `H_(2)SO_(4)` in 1 L solution = 98 g `H_(2)SO_(4)` in 1500 g solution = 98 g `H_(2)SO_(4)` in 1402 g water.
Also, in 109% oleum, 9 g `H_(2)O` is required to form 109 g pure `H_(2)SO_(4)` & so, to prepare 98 g `H_(2)SO_(4)`, water needed is 9/109 `xx 98 = 8.09 g.`
Total water needed = 1402 + 8.09 = 1410.09 g = 1410.09 mL