Asked in Elements and Compounds
What is the freezing point depression of an aqueous solution containing 25.0 grams of magnesium chloride?
June 05, 2007 10:58PM
You need to know the mass of the water (in kilograms) in the
solution to do this problem. Without that, you cannot solve for the
However, if you did know that, then you could use the following equation:
∆T = i * Kf * m
where ∆T is the change in the freezing point, "i" is the number of molecules or ions formed upon addition to the solvent, Kf is the freezing point depression constant, and "m" is the molality of the solution (this is where you need the amount of water).
The Kf of water for a freezing point depression is known and is equal to -1.858 K·kg/mol (or -1.858 °C molal-1)
The value of "i" has to do with what you add to the water. If you added sugar, a molecular compound the value of "i" is 1.0. If you add a ionic compound like NaCl, the value of "i" is 2.0 because for every 1 molecule of NaCl, you make 2 ions: one Na+ and one Cl- in water. For MgCl2, the value of "i" is thus 3.0 (for each MgCl2 you get one Mg2+ and two Cl- ions, so a total of 3 ions).
To find "m," the molality of a solution you need to know the number of moles of solute and the number of kilograms of solvent (m = moles/kg). The number of moles of solute here is found from the mass of the MgCl2 and its molar mass. However, you can't actually find the molality of the solution without knowing how many kilograms of water are in it.
The molar mass of MgCl2 is:
24.305 + 2*35.453 = 95.211 grams per mole.
So if you have grams, you divide by the molar mass to get moles:
25.0 ÷ 95.211 moles = 0.26257 moles.
So to find the freezing point depression, you put this all together like this:
∆T = 3.0 * -1.858 °C·kg/mol * 0.26257 mol ÷ (insert kilograms of water here)