Question

please help i dont understand it
1.What is the boiling point of a solution of 675 grams of ethylene glycol (C2H6O2) in 2.50 liters of water?

Answer 1

Answer:-  solution boiling point = 102.23 degree C (102 degree C with three sig figs).

Solution:- When a non volatile solute is added to a solvent then boiling point increases. Elevation in boiling point is directly proportional to the molality of the solution.

The equation is:

$\Delta T_b=i*k_b*m$

where, $\Delta T_b$ is the elevation in boiling point, i is the Van't hoff factor, $k_b$ is the molal elevation constant and m is the molality.

Value of i is 1 as ethylene glycol is a covalent molecule that does not break to give ions. $k_b$ for water is $\frac{0.512^0C}{m}$ .

We can calculate the molality from the given grams of ethylene glycol and liters of water as molality is moles of solute per kg of solvent.

Molar mass of ethylene glycol is 62 gram per mol and density of water is 1.00 kg per liter.

$2.50L(\frac{1kg}{1L})$

= 2.50 kg

Let's calculate the moles of ethylene glycol.

$675g(\frac{1mol}{62g})$

= 10.9 mol

molality of the solution = $\frac{10.9mol}{2.50kg}$

= 4.36m

Let's plug in the values in the equation we have on the top for elevation in boiling point.

$\Delta T_b=4.36m(\frac{0.512^0C}{m})$

= $2.23^0C$

Boiling point of pure water is 100 degree C. So, the boiling point of the solution = 100 + 2.23 = 102.23 degree C

(If we fix the three sig figs then it could be written as 102 degree C.)

Answer 2

Answer:

102 °C

Explanation:

The boiling point elevation is computed as follows:

ΔT = k*m*i

where ΔT is the difference between the solution's boiling point and water's boiling point, k is a constant (0.51 °C*kg/mol for water), m is the molality of the solution, and i is the van't Hoff constant (equal to 1 for ethylene glycol  because is a covalent molecule and it doesn't dissociate).

Molar mass of ethylene glycol: 62.07 g/mol

Moles of ethylene glycol: mass / molar mass = 675/62.07 = 10.87 mol

Density of water: 1 kg/L

Mass of water: density * volume = 1*2.5 = 2.5 kg

Molality of the solution: moles of solute / kg of solvent = 10.87/2.5 = 4.348 mol/kg

Then:

ΔT = 0.51*4.348*1 = 2.217 °C

Boiling point of water: 100 °C

Then, the boiling point of the solution is 100 + 2.217 ≈ 102 °C