Question

Some ethylene glycol, , is added to your car’s cooling system along with 5.0 kg of water.

a. If the freezing point of the water–glycol solution is −14.0 °C, what mass of must have been added?
b. What is the boiling point of the coolant mixture? Kb(H20) = 0.52 degrees celcius kg mol^-1.

Answer 1

a) 2.333 kg of glycol was added in car's cooling system along with 5.0 kg of water.

b) The boiling point of coolant mixture is 103.91°C.

Step-by-step explanation:

a)
`\Delta T_f=T-T_f`

`\Delta T_f=K_f* m`

where,

`T_f` = freezing point of solution

T = freezing point of solvent

`\Delta T_f` =depression in freezing point

`K_f` = freezing point constant

m = molality =
`\frac{moles}{\text{Mass of solvent(kg)}}`

we have :

Mass of glycol = x

Mass of solvent = 5.0kg

Molality of the solution ,m=
`(x)/(62 g/mol* 5.0 kg)`

`\Delta T_f=T-T_f`

`=0^oC-(-14.0)^oC=14^oC`

`K_f` =1.86°C/m ,

`14.0^oC=1.86^oC* (x)/(62 g/mol* 5.0 kg)`

x = 2,333.33 g

2,333.33 = 2.3333 kg ( g = 0.001 kg)

2.333 kg of glycol was added in car's cooling system along with 5.0 kg of water.

b)

`\Delta T_b=K_b* m`

where,

`\Delta T_b` =elevation in boiling point =

`K_b` = boiling point constant

m = molality

we have :

`K_b` =0.52°C/m ,

Molality of the solution ,m=
`(2,333.33 g)/(62 g/mol* 5.0 kg)=7.527 m`

`\Delta T_f=0.52^oC* 7.527 m`

`\Delta T_f=3.91^oC`

Boiling point of pure water = T = 100°C

Boiling point of solution =
`T_b`

`\Delta T_b=T_b-T`

`T_b=T+\Delta T_b=100^oC-3.91^oC=103.91^oC`

The boiling point of coolant mixture is 103.91°C.