Question

Calculate the freezing point of the coolant in an engine consisting of 4.23 kg of ethylene glycol (C2H6O2 ) and 5.04 kg of water. Kf for water is 1.86°C/m

A) -0.034 °C B) -2.52 °C C) 25.15 °C D) -25.2 °C E) 0 °C

Which of the following aqueous solutions has the highest freezing point (assume 100% dissociation for all soluble ionic compounds)?

A) 0.10m Mg(NO3 )2 (an electrolyte) B) 0.11m Na2SO3 (an electrolyte) C) 0.17m MgCO3 (an electrolyte) D) 0.18m NaCl (an electrolyte) E) 0.29m C6H12O6 (not an electrolyte)

Answer 1

1)

Lets calculate molality first

Molar mass of C2H6O2,

MM = 2*MM(C) + 6*MM(H) + 2*MM(O)

= 2*12.01 + 6*1.008 + 2*16.0

= 62.068 g/mol

mass(C2H6O2)= 4.23 kg

= 4230.0 g

number of mol of C2H6O2,

n = mass of C2H6O2/molar mass of C2H6O2

=(4230.0 g)/(62.068 g/mol)

= 68.15 mol

m(solvent)= 5.04 Kg

Molality,

m = number of mol / mass of solvent in Kg

=(68.15 mol)/(5.04 Kg)

= 13.52 molal

lets now calculate ΔTf

ΔTf = Kf*m

= 1.86*13.522036

= 25.150986 oC

This is decrease in freezing point

freezing point of pure liquid = 0.0 oC

So, new freezing point = 0 - 25.150986

= -25.2 oC

Answer: B

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