The Tale of Two Freezing Points
Imagine you are standing in a laboratory with two beakers in front of you. One contains a 4% aqueous solution of a mysterious substance X, and the other holds a 12% aqueous solution of another substance Y. You dip a thermometer into both and discover something fascinating: they freeze at the exact same temperature!
This simple observation is the key to unlocking the molecular weight of Y. Let's dive into the beautiful world of colligative properties to see how.
The Molality Connection
When two solutions share the same freezing point, it means their depression in freezing point, denoted by ΔTf, is identical. The formula for this depression is:
Since both are aqueous solutions, their solvent is water. This means the molal depression constant, Kf, is exactly the same for both. If we equate their ΔTf values, the Kf beautifully cancels out, leaving us with a profound realization: their molalities must be equal.
Decoding the Percentages
Now, we need to carefully extract the molality from the given percentages. A 4% aqueous solution by mass means that for every 100 g of the solution, there are 4 g of solute X.
Here is where many students make a classic silly mistake: they take the mass of the solvent as 100 g. But remember, the solvent mass is the total mass minus the solute mass! So, the mass of water is 100−4=96 g.
The molality of X is:
Similarly, for the 12% solution of Y, we have 12 g of solute in 88 g of water (100−12=88). Its molality is:
Crunching the Numbers
We know that mX=mY, and we are given that the molecular weight of X is A (MX=A). Let's set up our master equation:
A⋅964×1000=MY⋅8812×1000
The 1000s cancel out immediately. Simplifying the fractions, we get:
Now, it's just a matter of cross-multiplying to isolate MY:
MY=223⋅24A=2272A=1136A
When we calculate 1136, we get approximately 3.27. So, MY≈3.27A.
Looking at our options, the closest integer multiple is 3A. And just like that, by understanding the physical reality behind the equations, we've cracked the code!