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 -10.°C? (The density of ethylene glycol is 1.11 g/mL. Assume the density of water at -10.°C is 1.00 g/mL.)

Answer 1

Answer:

Around 2.0 L of ethylene glycol needs to be added to the car radiator

Explanation:

The depression in freezing point ΔTf of a solution is directly proportional to its molality (m), i.e.

$\Delta T_{f}= T_{f}^{0}-T_{f}=i*K_{f}*m$

From the given information:

$T_{f}$ = freezing pt of solution = -10.0 C

$T_{f}^{0}$ = freezing pt of pure solvent = 0 C

Kf = freezing pt depression constant = 1.86 C/m

i = 1 for ethylene glycol antifreeze

$[0-(-10.0)] C= 1*(1.86 C/m) *( m)\\\\m = 5.38$

$Molality = \frac{moles\ of\ solute}{kg\ solvent} \\\\Therefore,\ moles of antifreeze = molality* mass\ of\ water$

$Molality = \frac{moles\ of\ solute}{kg\ solvent} \\\\Therefore,\ moles of ethylene glycol = molality* mass\ of\ water$

Volume of water = 6.50 L = 6500 ml

Density of water = 1.00 g/ml

Therefore mass of water = $density * volume = 1.00g/ml*6500ml = 6500g = 6.50kg$

$moles\ of\ ethylene glycol= 5.38moles/kg*6.50kg = 34.9 moles$

Molar mass of ethylene glycol = 62 g/mol

Mass of ethylene glycol needed = $molar\ mass* moles = 62g/mol*34.9moles=2163.8g$

Density of ethylene = 1.11 g/ml

Therefore, volume needed = $\frac{mass}{density} =\frac{2163.8g}{1.11g/ml} =1949ml$

Answer 2

 16 = m x 1.86 
m = 8.60 = moles solute / 6.50 Kg 

moles solute = 55.9 

mass solute = 55.9 x 62.068 g/mol=3470 g 

V = 3479/ 1.11 =3126 mL= 3.13 L 

delta T = 8.60 x 0.512 =4.40 
boling point = 104.4 °C

Hope this helps.