Question

Ethylene glycol (C2H6O2) is molecular compound that is used in many commercial anti-freezes. An aqueous solution of ethylene glycol is used in vehicle radiators to prevent the water in the radiator from freezing by lowering its freezing point. Calculate the freezing point of a solution of 400g of ethylene glycol in 500g of water.
Kf (H2O) = -1.86 o C/m

Answer 1

Hope it helps you

Please mark as brainliest

Answer 2

Answer:

The freezing point of this solution will be - 13.32°C.

Explanation:

The freezing point depression of the solution is given by the equation as follows:

Δ$T_{f} =k_{f} m$

Where we have Δ$T_{f}$ = the lowering of the freezing point

$k_{f}$ = cryoscopic constant = 1.86 K Kg/mol and m = the molality

First, we will find out the moles of ethylene glycol solution in water by the following given formula:

No. of moles of ethylene glycol = mass of ethylene glycol / molar mass of ethylene glycol

$C_{2} H_{6} O_{2}$ = 2×12g + 6×1g + 2×16g = 62g

No. of moles = $\frac{400g}{62g} = 6.45 moles$

Now, we find out the molality,

molality = no of moles of the given solute / the total mass of given solution in Kg

Total mass of the solution = 400g + 500g = 900g = 0.9 Kg

Molality = $\frac{6.45}{0.9} = 7.16 m$

Now, we have the freezing point lowering as:

Δ$T_{f} = k_{f} m$

Δ$T_{f}$ = 1.86 × 7.16 = 13.3176

Now we know that the water freezing point is 0° C

Therefore the freezing point = 0°C - 13.3176°C = -13.3176°C

Hence we get the final answer to be - 13.32°C.