If at a particular temperature, the density of `18MH_(2)SO_(4)` is `1.8g cm^(-3)`, calculate (a) molality, (b) `%` concentrating by weight of solute and solvent (c) mole fraction of water and `H_(2)SO_(4)`.

Answer

`0.81,95.87`

Explanation

18 `M H_(2)SO_(4)` means 18 moles of `H_(2)SO_(4)` are present in 1000 mL of solution .
Mass of solution ` = 100 xx 1.84 = 1840 g`
Mass of `H_(2)SO_(4) = 18 xx 98 = 1746 g`
Mass of water ` = 1840-176 = 76 g`
Moles of water ` = (76)/(18) = 4.22`
Mole fraction of `H_(2)SO_(4) = (18)/(18+4.22) = 0.81`
Mass % `H_(2) SO_(4) = (1764)/(1840) xx 100 = 0.02`