Question

Calculate the mole fraction and molality of HNO₃ in solution contaning 12.2%HNO₃ (Given atomic mases:H=1, N=13,O=16)

Answer 1

%purity = 12.2 %

∴mass of solute = 12.2 g.

mass of solvent = 100 – 12.2

= 87.8 g

`:."Mole of solute"=12.2/63=0.193`

`:."Mole of solvent"=87.7/18=4.877`

`:."mole fraction of solute"=0.193/(0.193+4.877)`

`=0.193/5.07`

= 0.038

Mole fraction of HNO₃ = 0.038

`"Molality of HNO"_3 = "Mass of HNO"_3/("Molar mass of HNO"_3 xx "mass of solvent")`

`"molality"=W_2/M_2.(1000)/W_1`

`=12.2/63xx1000/87.8`

`=12200/5531.4`

= 2.2056

∴ Molality = 2.2056 mol/kg