Depression of Freezing Point

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. In many cold places, in winter, people spread salt on icy roads. Salt, on ice! Salt does not heat anything. And yet the ice melts, even when the air is below zero degrees Celsius. How can that be? Strange, right? Today: the depression of freezing point. It comes up often in JEE, so let's begin.

0:30Why a solution freezes lower

So, here is the graph from the last lesson: vapour pressure, small p, going up, and temperature, capital T, along the bottom. The right side was boiling; today, the left side. The grey curve is the solid, dark blue is the pure solvent, and green is the solution. With me? Now, the pure solvent freezes where its blue curve meets the grey solid curve. That temperature is T f nought: capital T, with a small f below and a little circle on top. The circle means the pure solvent. Okay? Next, the green curve. It lies below the blue one, because a non-volatile solute lowers the vapour pressure, remember? Where does the green curve meet the grey one: at a higher temperature, or a lower one? Lower, yes, exactly! That point is T f, with no circle: the freezing point of the solution. So the solution freezes below the pure solvent. Right? The gap between them is the small orange double arrow on the left. We call it delta T f. The little triangle is the Greek letter delta, and it means a change. Delta T f equals T f nought minus T f: the depression of freezing point. Clear? Puzzle solved! The salt lowers the freezing point of water, so the ice melts even below zero degrees Celsius, the small circle and capital C on screen. Got it?

1:49How big is the drop?

So, how big is this drop? Does it depend on what the solute is? Sugar, urea, glycol? No, it does not! Depression of freezing point is a colligative property: it depends only on the number of solute particles, not on their nature. So, delta T f equals K f times m. That's it! Here m is the molality: moles of solute per kilogram of solvent. Oh no, careful! Molality, not molarity. Per kilogram of solvent, not per litre of solution. Okay? And K f, capital K with a small f below, is the cryoscopic constant. Its unit is kelvin kilogram per mole: the K, the k g, and mol to the power minus one on screen. It depends only on the solvent. With me? Remember three values. Water: one point eight six. Benzene: five point one two. Camphor: about forty, the wavy equals sign means approximately. So in camphor, even a little solute gives a big drop, easy to measure. Right? And the conditions: a dilute solution, a non-volatile solute, and a solute that neither splits up nor joins up, so no dissociation or association. Those cases come later, with the van 't Hoff factor i. Got it?

3:07Finding the molar mass

Now, finding a molar mass. Molality is w B over M B, the moles of solute, divided by w A over one thousand. Here w B, the w with a small B below, is the mass of solute, and w A the mass of solvent, both in grams; M B is the molar mass of the solute. So why divide w A by one thousand? Yes, exactly! Grams to kilograms, because molality is per kilogram. Now put this m into delta T f equals K f times m, and rearrange. M B equals one thousand times K f times w B, over delta T f times w A, okay? Measure the drop, and you get the molar mass. Clear?

3:49Where does Kf come from?

Where does K f come from? K f equals R times T f nought squared, times M A, all over one thousand times delta fus H, with fus written small below. R is the gas constant. T f nought, with the little circle, is the freezing point of the pure solvent, in kelvin, and the small two outside the bracket means squared; M A is its molar mass, and delta fus H is its enthalpy of fusion per mole. Only the solvent, no solute anywhere! Okay? For water, the enthalpy of fusion is six point zero one kilojoules per mole. Oh no, careful! R is in joules, so use six thousand and ten joules per mole. Right? Now: eight point three one four, times two hundred seventy three point one five squared, times eighteen, over one thousand times six thousand and ten; the cross means times. One point eight six kelvin kilogram per mole, our number for water! Got it?

4:46Only the solvent freezes

One last idea. When a solution freezes, what turns solid: the solvent, the solute, or both? Only the solvent! In water, the ice that forms is pure water; the solute stays in the liquid. Okay? So the liquid left behind has the same solute in less solvent: more concentrated, so its freezing point drops further. Example two depends on this. With me?

5:11Example 1: freezing point of ethylene glycol in water

Now, example one. Thirty one grams of ethylene glycol, capital M equals sixty two, its molar mass, is dissolved in five hundred grams of water. K f is one point eight six. Find the freezing point. Pause the video, and try it first. Done? Okay, let's check. Moles of glycol, n B, the n with a small B below: thirty one over sixty two, that is zero point five mole. Right? Five hundred grams of water is zero point five kilogram. So the molality is zero point five mole over zero point five kilogram: one mole per kilogram. Okay? So, delta T f equals one point eight six times one: one point eight six kelvin. So the freezing point is one point eight six, right? No, no, careful! One point eight six is the drop, not the freezing point. Take it away from pure water's freezing point, two hundred seventy three point one five kelvin: two hundred seventy one point two nine kelvin. That is minus one point eight six degrees Celsius, since a drop of one kelvin is a drop of one degree. Clear?

6:21Example 2: how much ice separates at -3.72 °C

Now a harder one, JEE level. Zero point two five mole of ethylene glycol is dissolved in two hundred fifty grams of water. The solution is cooled to minus three point seven two degrees Celsius. How much ice separates? Pause and try. Hint: only the water freezes. Done? Okay, let's check. At first, the molality is zero point two five mole over zero point two five kilogram: one. So it starts to freeze at minus one point eight six degrees Celsius. But we cooled it lower, to minus three point seven two, so ice has formed. With me? The liquid left behind is just at its freezing point, minus three point seven two. So what must its molality be? Delta T f is three point seven two kelvin, so m equals three point seven two over one point eight six: two moles per kilogram. Yes! The liquid is now twice as concentrated. Okay? The glycol does not freeze, so all zero point two five mole is still in the liquid. So the water left is zero point two five over two: zero point one two five kilogram, that is one hundred twenty five grams. Right? So the ice is two hundred fifty minus one hundred twenty five: one hundred twenty five grams. Half the water froze, and only the water! That's it!

7:40Quick revision: delta Tf = Kf m and the molar mass

So, quick revision. Delta T f equals K f times m, with molality, for a dilute solution. For molar mass: M B equals one thousand K f w B, over delta T f w A, okay? And only the solvent freezes out, so the rest gets more concentrated. Now go and try the practice problems on this page. All the best!