Question

Fill in the blanks in the table. Aqueous solutions are assumed.

Compound	Molality	Weight percent	Mole Fraction
NaI	0.15
C2H5OH		5.0
C12H22O11	0.15

Answer 1

NaI: Molality= 0.15

Amount of NaI present in 1000 gm water

= 0.15 moles

= 0.15 moles *149.89 g/mol

= 22.4835 g

1000 gm water = 1000/18 = 55.55 moles of water

So, moles fraction of NaI = 0.15/(0.15+55.55) = 2.69*10^-3

Weight %

= [weight of ethanol in 1000 g water /(weight of total solution) ] * 100

= [22.4835/(1000 + 22.4835) ] * 100

= 2.20 %

C₂H₅OH : Weight percent of solution = 5 %

So 1000 gm solution contains 50 g ethanol.

(1000-50) or 950 g water is mixed with 50 g ethanol.

50 g ethanol = 50g/(46 g/mol) = 1.087 moles of ethanol

950 g water contains 1.087 moles of ethanol

1000 g water contains 1.087*1000/950 or 1.144 moles of ethanol.

Molality = 1.144

mole fraction of ethanol= 1.144/(55.55+1.144) = 0.020

C₁₂H₂₂O₁₁ : Molality= 0.15

Amount of C₁₂H₂₂O₁₁ present in 1000 gm water

= 0.15 moles

= 0.15 moles *342 g/mol

= 51.3 g

1000 gm water = 1000/18 = 55.55 moles of water

So, moles fraction of NaI = 0.15 /(0.15+55.55)

= 2.69*10^-3

Weight %

= [weight of C₁₂H₂₂O₁₁ in 1000 g water /(weight of total solution) ] * 100

= [51.3/(1000 + 51.3) ] * 100

= 4.88 %