Question

What is the freezing point of a solution that contains 10.0 g of glucose (C6H12O6 ) (a nonelectrolyte) in 100.0g of H2O? Kf for water is 1.86°C/m.

Which one of the following substances is expected to have the lowest melting point? A) KI B) CsI C) LiI D) NaI E) RbI

Answer 1

1)

Lets calculate molality first

Molar mass of C6H12O6,

MM = 6*MM(C) + 12*MM(H) + 6*MM(O)

= 6*12.01 + 12*1.008 + 6*16.0

= 180.156 g/mol

mass(C6H12O6)= 10 g

number of mol of C6H12O6,

n = mass of C6H12O6/molar mass of C6H12O6

=(10.0 g)/(180.156 g/mol)

= 5.551*10^-2 mol

m(solvent)= 100 g

= 0.1 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(5.551*10^-2 mol)/(0.1 Kg)

= 0.5551 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*0.555074

= 1.03 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 1.03

= -1.03 oC

Answer: -1.03 oC

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