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