Question

How many liters of the antifreeze ethylene glycol [CH2(OH)CH2(OH)] would you add to a car radiator containing 6.50 L of water if the coldest winter temperature in your area is –20ºC? Calculate the boiling point of this water-ethylene glycol mixture. (Te density of ethylene glycol is 1.11 g/mL.) Kf = 1.86 ºC /m

Answer 1

Answer : The volume of ethylene glycol is, 3.92 liters.

The boiling point of a solution is, $105.6^oC$

Explanation :

First we have to calculate the molality of ethylene glycol.

Formula used :  

$\Delta T_f=i\times K_f\times m$

where,

$\Delta T_f$ = change in freezing point = $20^oC$

i = Van't Hoff factor = 1   (for non-electrolyte)

$K_f$ = freezing point constant = $1.86^oC/m$

m = molality of ethylene glycol = ?

Now put all the given values in the above formula, we get:

$20^oC=1\times 1.86^oC/m\times m$

$m=10.8mol/kg$

Now we have to calculate the mass of water.

$\text{Mass of water}=\text{Volume of water}\times \text{Density of water}$

$\text{Mass of water}=6.50L\times 1kg/L=6.50kg$

Now we have to calculate the moles of ethylene glycol.

Moles of ethylene glycol = Molality × Mass of water

Moles of ethylene glycol = 10.8 mol/kg × 6.50kg = 70.2 mole

Now we have to calculate the mass of ethylene glycol.

Mass of ethylene glycol = Moles of ethylene glycol × Molar mass of ethylene glycol

Mass of ethylene glycol = 70.2 mole × 62.07 g/mole =4357.3 g

Now we have to calculate the volume of ethylene glycol.

$\text{Volume of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Density of ethylene glycol}}$

$\text{Volume of ethylene glycol}=\frac{4357.3g}{1.11g/mL}=3925.49mL=3.92L$

The volume of ethylene glycol is, 3.92 liters.

Now we have to calculate the boiling point of solution.

Formula used for Elevation in boiling point :

$\Delta T_b=i\times k_b\times m\\\\T_b-T^o_b=i\times k_b\times m$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of pure water = $100^oC$

$k_b$ = boiling point constant  = $0.52^oC/m$

m = molality  = 10.8 mol/kg

i = Van't Hoff factor = 1   (for non-electrolyte)

Now put all the given values in the above formula, we get the boiling point of a solution.

$T_b-100^oC=1\times 0.52^oC/m\times 10.8mol/kg$

$T_b=105.6^oC$

Therefore, the boiling point of a solution is, $105.6^oC$

Answer 2

The required volume of the antifreeze ethylene glycol in the car radiator is 3.92 L.

How to calculate the volume of antifreeze?

The volume can be calculated by the use freezing point depression formula,

$V = \dfrac m\rho$............1

Where,

$V$ - volume

$m$ - mass

$\rho$ - density = 1.11 g/mL

First, calculate the molality of ethylene glycol:

$m = \dfrac {\Delta t }{k_f \times i}$

Where,

$m$ - molality

$\Delta T$ - change in freezing point = 20°C

$k_f$ - freezing point constant  = 1.86 C/m

$i$  -  Van't Hoff factor = 1  

Put the values in the formula, we get:

m = 10.8 mol/kg

Now calculate the moles of Ethylene glycol,

n = 10.8 mol/kg  x 6.5 kg

n = 70.2 mole

Now calculate the mass,

w = 70.2 mole x  62.07

w = 4357.3 g

Put the values in equation 1,

$V =\dfrac {4357.3}{ 1.11}\\\\V = 3.92 \rm \ L$

Therefore, the required volume of the antifreeze ethylene glycol in the car radiator is 3.92 L.

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