Question

The amount of ice that will separate out on cooling a solution containing 50 gram of Ethylene glycol in 200 gram water to -9.3⁰C

(K_f=1.86K/molal)

Answer 1

We know that freezing point of pure water is 0^oC. Using the formula , ΔT_f = K_f m
where ‘m’ is the molality . First we will calculate the value of m --->
m = no.of moles of solute / wt. of solvent in kg

molar mass of ethylene glycol ( C₂H₆O₂ ) = 62 g/mol
no.of moles of solute = 50 / 62 = 0.806 moles

Now, m= 0.806 / (200 * 10^-3 ) = 4.03 molal
So,
ΔT_f = K_f m = 1.86 * 4.03 = 7.49 = 7.5 ^oC ( approx )

ΔT_{f =} T_f^o – T_f
7.5 = 0 – T_f
So, T_f = -7.5 ^oC
Notice that we are cooling the solution to the temperature -9.3^oC which is lower than -7.5 ^oC . This implies that molality of the solution will increase because some of the water turns into ice , thus leaving less water in the solution.

The no. of moles of ethylene glycol (solute) will remain unchanged .

So, we can write , ΔT_{f cool}= K_f m_new
m_new ---> molality of the solution at this new temp.
T_f^o – T_f = 0- (-9.3) = 9.3 = 1.86 * (0.806 / wt. of solvent in Kg)


wt. of solvent = 0.1612 kg or 161.2 g
this is the amount of water at -9.3 ^oC . So, we can say that the amount of ice that will seperate out is :


mass of ice = 200g -161.2g = 38.8 g