Question

What is the freezing point of a 50% by volume ethanol solution? Assume water is the solvent. (Use data from Table 12.7 in the textbook.)

Solvent	Normal Freezing Point (°C)	Kf (°C m−1)	Normal Boiling Point (°C)	Kb (°C m−1)
Benzene (C6H6)	5.5	5.12	80.1	2.53
Carbon tetrachloride (CCl4)	−22.9	29.9	76.7	5.03
Chloroform (CHCl3)	−63.5	4.70	61.2	3.63
Ethanol (C2H5OH)	−114.1	1.99	78.3	1.22
Diethyl ether (C4H10O)	−116.3	1.79	34.6	2.02
Water (H2O)	0.00	1.86	100.0	0.512

For water:

Kb = 0.512 °C m−1

Answer 1

Given, 50 % by volume ethanol solution,

It means 50 mL ethanol in 100 mL solution.

Thus, the volume of solvent(water) = 100 mL solution - 50 mL ethanol = 50 mL of water

We know,

The density of ethanol = 0.789 g/mL

The density of water = 1.00 g/mL

Converting the volumes of ethanol and water to mass by using the density values,

= 50 mL ethanol x ( 0.789 g / 1 mL)

= 39.45 g of ethanol

Also,

= 50 mL water x ( 1.00 g / 1 mL)

= 50 g of water

= 0.050 kg of water

Now,Calculating the number of moles of ethanol from the mass,

= 39.45 g of ethanol x ( 1 mol /46.07 g)

= 0.8563 mol of ethanol.

Now, Calculating the molality of ethanol solution,

We know, Molality(m) = Number of moles / Kg of solvent

m = 0.8563 mol of ethanol / 0.050 kg of water

m = 17.13 m

Now, We know the formulae,

T_f = K_f x m

also, we know,

T_f = F.P(solvent) - F.P of solution

Now,

T_f = 1.86 ^oC/m x 17.13 m

T_f = 31.85 ^oC

Now,

T_f = F.P(solvent) - F.P of solution

31.85 ^oC = 0.0 ^oC - F.P of solution

F.P of solution = -31.85 ^oC

Thus, the freezing point of the 50 % by volume ethanol solution is -31.9 ^oC Or -32 ^oC