Molarity of `H_(2)SO_(4)` is 0.8 M and its density is `1.06 g cm^(-3)`. What will be concentration of the solution in terms of molality and mole fraction ?

Explanation

Molarity of `H_(2)SO_(4)=0.8` means 0.8 mole `H_(2)SO_(4)` are present in 1 L of the solution
`"0.8 mole "H_(2)SO_(4)=0.8xx98 g = 78.4g," 1 L "H_(2)SO_(4) " solution "=1000xx1.06g=1060 g`
`therefore" Mass solvent (water )"=1060-78.4 g = 981.6g=(981.6)/(18)" moles"="54.53 moles"`
Thus, 981.6 g of water contain 0.8 mole of `H_(2)SO_(4)`.
`therefore" Molality "=("0.8 mol")/("981.6 g")xx"1000 g kg"^(-1)="0.815 mol kg"^(-1)`
Further, the solution contains 0.8 mole of solute in 54.53 moles of solvent.
`therefore" Mole fraction of solute"=(0.8)/(0.8+54.53)=0.014`