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.)

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.

Answer Link

Annainn Annainn · Answer 2 · 2019-09-21T03:16:07-04:00

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 = \(\frac{mass}{density} =\frac{2163.8g}{1.11g/ml} =1949ml\)