Concentration Terms and their Interconversion

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take one bottle of concentrated sulphuric acid. One student says it is eighteen point four molar. Another says it is five hundred molal. Eighteen, and five hundred, for the very same bottle? Can both of them be right? Hmm. Hold that thought. By the end of this lesson, you will settle it yourself. Concentration terms come up often in JEE, and the rest of this unit is built on them. So, let's begin.

0:34Percentages and parts per million

So, first the names. The substance we dissolve is the solute, called B, and the liquid it dissolves in is the solvent, called A, okay? Every formula today uses these two letters. Now the simplest measure: mass percentage. Mass percentage equals the mass of B, over the mass of the solution, times one hundred. The percent sign just means per hundred. Right? Quick check. Five grams of common salt is dissolved in forty five grams of water. So the mass percentage is five over forty five, times one hundred, right? No, no, careful! The bottom is the mass of the whole solution: five plus forty five, that is fifty grams. The small g on screen is grams, and N a C l is the salt. So, five over fifty, times one hundred: ten percent. With me? For a liquid in a liquid, we often use volume percentage instead: the same idea, with volumes. And mass by volume means grams of solute in one hundred millilitres, the m L on screen. Clear? Now, when the solute is only a trace, percentages become tiny. So we use parts per million, p p m: the mass of B over the mass of the solution, times ten to the power six, that is one million. Got it?

1:53Counting moles: mole fraction, molarity, molality

Now let's count moles. n with the subscript B is the moles of solute, and n with the subscript A is the moles of solvent. The mole fraction of B, x with the subscript B, equals n B over n A plus n B: its share of all the moles. And the two fractions together make one: x A plus x B equals one. Okay? Next, molarity, which we write as capital M, okay? It is the moles of solute, n B, over the volume of the solution, V with the subscript solution, in litres. The capital L in brackets means litres. Right? Now molality, written small m. Moles of solute, divided by a mass in kilograms. But the mass of what? The solution, or the solvent? The solvent, yes! Molality, small m, equals n B over the mass of the solvent, in kilograms, the k g in brackets. Clear? And this is the most common trap here. Molarity uses the volume of the whole solution, but molality uses the mass of the solvent only, not of the solution. Don't fall for it! With me?

3:05Warm the solution: which terms change?

Alright, a quick question. Take four terms: mass percentage, mole fraction, molality and molarity. You warm the solution a little; nothing is added, nothing escapes. Which of these terms change? Mass percentage, mole fraction and molality do not change with temperature, the capital T on screen. Why? They are built from masses, and masses do not change when you warm something. Okay? But molarity does change. It divides by the volume of the solution, and the volume of a liquid changes with temperature. So the same solution gives a slightly different molarity when it is warmer. Got it?

3:45Interconversion: think of one litre

Now, converting one term into another. Two more symbols. d is the density, in grams per millilitre, that is g m L to the power minus one. M with the subscript B and M with the subscript A are the molar masses of the solute and the solvent. Careful: capital M alone is molarity; with a subscript, it is a molar mass. Right? Now, don't just memorise. Take one litre of the solution, that is one thousand millilitres. What is its mass in grams? One thousand times d, exactly! Of that, the solute is M moles, so M times M B grams. The rest is solvent: one thousand d, minus M times M B, in grams. Divide the moles by that mass in kilograms, and you get the formula: molality, small m, equals one thousand M, over one thousand d minus M times M B, okay? Next, from mass percentage, w, to molarity. Again one litre: its mass is one thousand d grams, and w percent of it is solute, that is ten d w grams. Divide by the molar mass: capital M equals ten d w over M B, with me? And from molality to mole fraction: x B equals m M A, over one thousand plus m M A, where M A is the molar mass of the solvent. Clear? One last point. For a dilute solution in water, m is approximately equal to capital M; the wavy equals sign means approximately. But for a concentrated solution, they differ a lot. Remember this: our puzzle is hiding right here.

5:29Example 1: molality of 2 M NaOH from its density

Now let's try one. Two molar sodium hydroxide, N a O H, has a density of one point zero eight grams per millilitre. Find its molality. Pause the video, and try it first. Done? Okay, let's check. Take one litre of the solution. Its mass is one thousand times one point zero eight: one thousand and eighty grams. Now, how much of this is water? Two moles of N a O H, at forty grams per mole, is eighty grams. So the water is one thousand and eighty minus eighty: exactly one thousand grams. Right? So, molality is two moles over one kilogram of water: two point zero moles per kilogram. That's it! And our formula agrees: one thousand times two, over one thousand and eighty minus eighty, gives two. Here molarity and molality came out equal, two and two. But that is not always true, as the next example shows.

6:29Example 2: molarity, molality, mole fraction of 98% H2SO4

Now our puzzle question. Concentrated sulphuric acid, H two S O four, is ninety eight percent by mass; the w slash w means mass by mass. Its density is one point eight four grams per millilitre. Find its molarity, its molality, and the mole fraction of the acid. Pause the video, and try it. Done? Okay, let's check. Molarity first, using ten d w over M B, our formula from before. Ten times one point eight four times ninety eight, over ninety eight, the molar mass of the acid. The ninety eights cancel: eighteen point four molar. Okay? Now molality. Eighteen point four moles in every litre, so the molality is also about eighteen point four, right? No, no, careful! That shortcut works only for dilute solutions. Take one thousand grams of this acid. Ninety eight percent of it, nine hundred and eighty grams, is the acid itself: ten moles. And only twenty grams is water: one point one one moles. With me? So molality is ten moles over the mass of the solvent, just twenty grams, that is zero point zero two zero kilograms. Five hundred molal! And there is our puzzle, solved. Same bottle: eighteen point four molar, but five hundred molal, because there is so little solvent. Both students were right! Finally, the mole fraction: ten over ten plus one point one one, that is ten over eleven point one one. Zero point nine zero. Got it?

8:10Quick revision: molarity vs molality, one-litre trick

So, quick revision. Molarity is per litre of solution; molality is per kilogram of solvent. Molarity changes with temperature; molality does not. And to convert, take one litre: its mass is one thousand d grams. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!