Ethylene glycol (C2H6O2) is used as an antifreeze in cars. If 400 g of ethylene glycol is added to 4.00 kg of water, what is the molality? Calculate how much the freezing point of water will be lowered. The freezing-point depression constant for water is Kf = –1.86°C/m.

Remember that molality is generally the standard for calculating the concentration, between molarity and molality. In this question we have to apply molality, given the units as moles and kilograms.

We can start by receiving the molality of this solution through the following formula...

[tex]\mathrm{m}=\mathrm{(mass}\mathrm{/molar\:mass)} * (1000/\mathrm{mass\:of\:solvent)},[/tex]

[tex]\mathrm{m} =\mathrm{(400.0 g / 62.07 g/mol) * (1000 / 4000.0 g)},[/tex]

[tex]\mathrm{m} = (400/62.07) * (1000 / 4000) = \left(\frac{400}{62.07}\right)\left(\frac{1000}{4000}\right) = \frac{1000}{620.7} = 1.61108\dots\mathrm{m}[/tex]

Now we can determine how much the freezing point of water will be lowered, given that the freezing-point depression constant for water is Kf= –1.86°C/m. Remember that " Kf " is the final temperature, so to determine how much the temperature of the water will decrease, we can note the change in temperature...

[tex]\mathrm{change\:in\:temperature} = ( - 1.86\mathrm{C/m})(1.61\mathrm{m}) = - 1.86 * 1.61 = -2.9946[/tex]

[tex]-2.9946 = \mathrm{About\:-3C}[/tex]

Hence the freezing point of water will be lowered 3°C.