Question

The radiator of a motor car holds 10 liter of water if the vehicle is to be used in gulmarg where the lowest temperature is -10 degree celsius then how much ethylene glycol is to be added to the water filled in the radiator

Answer 1

Answer:

excusme is your question right or is your this only

Answer 2

Given: Amount of water held by the radiator = 10 liters

The lowest temperature in Gulmarg = -10°C

To Find: Amount of ethylene glycol to be added

Solution:

Let the mass of ethylene glycol be a

Some data that we should already know

Density of water = 1 g/ml

The freezing point of water = 0°C

Molal freezing point depression of water (kf) = 1.86 °C/m

Amount of water in ml = Amount of water in liters x 1000

Amount of water in ml = 10 x 1000

                                      = 10000 ml

Mass = Volume x Density

Mass of water = 10000 x 1

                        = 10000g

ΔTf = i x Kf x m

where ΔTf is the lowering of freezing point, i is the Vant Hoff factor and m is the molality of the solution.

Here we need to lower the freezing point to -10°C so that the water in the radiator does not freeze.

So, the change in freezing point will be 10

Here i will be 1 since we are not taking any salt or ionic compound

Molality (m) = Moles of solute / Mass of solution in kg

Moles of solute = Mass of solute given / Molar mass of solute

Here solute is ethylene glycol with formula C₂H₆O₂

The molar mass of ethylene glycol = 2x the Molar Mass of C + 6x the Molar mass of H + 2x the Molar mass of O

The molar mass of ethylene glycol = 2x12 + 6x1 + 2x16

                                                          = 62units

Moles of ethylene glycol = a/62

m = $\frac{a}{62}\frac{1000}{10000}$

m = $\frac{a}{620}$

10 = 1 x  1.86 x  $\frac{a}{620}$

a = 10000/3

a = 333.34

Therefore, the mass of ethylene glycol required is 333.34g