Question

How many moles of ethylene glycol must be added to 1 kg of water to make a solution with a freezing point of -2.4°C? The freezing point depression constant for water is 1.86°C•kg/mol. What is the molality of the solution?

Answer 1

Answer: 1.29 moles of ethylene glycol must be added to 1 kg of water to make a solution with a freezing point of -2.4°C. Molality of solution is 1.29 m.

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(0-(-2.4))^0C=2.4^0C$ = Depression in freezing point

i= vant hoff factor (for non electrolyte , i = 1)

$K_f$ = freezing point constant for water= $1.86^0C/kgmol$

m= molality =$\frac{\text{moles of solute}}{\text{weight of solvent in kg}}$

m= molality =$\frac{x}{1kg}$

$2.4^0C=1\times 1.86^0C/kgmol\times \frac{x}{1kg}$

$x=1.29$

Molality = $\frac{\text{moles of solute}}{\text {weight of solvent in kg}}=\frac{1.29mol}{1kg}=1.29m$

Thus 1.29 moles of ethylene glycol must be added to 1 kg of water to make a solution with a freezing point of -2.4°C. Molality of solution is 1.29 m.

Answer 2

Answer:

1) C (m=T f/Kf)

2) 2.4 C

3) 1.3 m

4) 1.3 moles

Explanation:

This is correct on ed