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

Answer:

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.

Explanation:

a) $\Delta T_f=T-T_f$

$\Delta T_f=K_f\times m$

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= $\frac{x}{62 g/mol\times 5.0 kg}$

$\Delta T_f=T-T_f$

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

$K_f$ =1.86°C/m ,

$14.0^oC=1.86^oC\times \frac{x}{62 g/mol\times 5.0 kg}$

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)

$\Delta T_b=K_b\times m$

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= $\frac{2,333.33 g}{62 g/mol\times 5.0 kg}=7.527 m$

$\Delta T_f=0.52^oC\times 7.527 m$

$\Delta T_f=3.91^oC$

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.