Converting Between Concentration Terms

Transcript

0:00Introduction

Hello everyone! Here is one small puzzle before we start. Take commercial sulphuric acid, ninety eight percent by mass. Its molarity is about eighteen point four. And its molality? About five hundred! Same bottle, same acid. Eighteen, and five hundred? Wait, how can one solution give two numbers so far apart? Is the book wrong? No. And by the end of this video, you will calculate both numbers yourself. You see, JEE loves these conversion questions, because one small slip gives a completely wrong answer. So, let's begin.

0:40One solution, three ways to count

So, first, let us see what each term is really counting. Molarity, capital M, is the moles of solute in one litre of solution, okay? Molality, small m, is the moles of solute in one kilogram of solvent. Not solution, solvent! And capital M and small m sound the same, so in the exam, write them carefully. Clear? And mass percent is the grams of solute in one hundred grams of solution. Now think. One term talks about litres of solution, one about kilograms of solvent, and one about grams of solution. So, to go from one to another, what extra information do we need? Yes, two bridges! Density, the letter d, turns a volume into a mass. And molar mass turns grams into moles. With these two, every conversion becomes possible. Got it?

1:35Mass percent to molarity: take 1 L

Alright, let's build the first formula ourselves. We know the mass percent and the density, and we want the molarity. Molarity talks about one litre of solution. So, how much solution should we take? Exactly, one litre! One litre is one thousand millilitres, and each millilitre weighs d grams. So the whole solution weighs one thousand d grams, okay? Now, how much of that is solute? The mass percent, written w by w percent on screen, tells us the grams of solute in every hundred grams. So, take the mass percent by one hundred, times one thousand d. That is simply ten, times the mass percent, times d, in grams. Divide by the molar mass of the solute, M sub B, to get the moles. These moles sit in one litre, so that is the molarity directly. Capital M equals ten, times the mass percent, times d, all divided by M sub B, okay? Really simple.

2:37Molarity to molality: remove the solute

Next, from molarity to molality. Again we take one litre of solution, and it has capital M moles of solute, with me? Molality needs the mass of the solvent. So, one litre of solution means one kilogram of water, right? No, no, careful! This is the JEE trap. One litre is the volume of the whole solution, solvent and solute together. Its mass is one thousand d grams, and the solute is also hiding inside it. So, the solution weighs one thousand d grams. The solute weighs capital M times M sub B grams, that is, moles times molar mass. Subtract, and what is left is the solvent: one thousand d, minus capital M times M sub B, in grams. Now divide the moles by the solvent mass in kilograms, and that brings in a factor of one thousand. Small m equals one thousand capital M, over one thousand d, minus capital M times M sub B, okay? Write this down. Got it?

3:39Mole fraction to molality

Now, one more conversion: from mole fraction to molality. If the mole fraction of the solute is x sub B, then for every x sub B moles of solute, there are one minus x sub B moles of solvent, okay? Multiply by the molar mass of the solvent, M sub A, and we get the solvent's mass in grams. Moles of solute over kilograms of solvent, and again the factor of one thousand. Small m equals one thousand x sub B, over one minus x sub B, times M sub A, okay? Now tell me, do we need the density here? No! See, neither mole fraction nor molality talks about volume, so density never enters. Clear?

4:22Topper's trick: never memorise blindly

Now, three formulas. Do we have to mug them all up? Toppers don't. Here is their trick. Step one: assume a convenient amount. One litre of solution goes with molarity, one kilogram of solvent with molality, and one hundred grams of solution with mass percent. Step two: find the moles and the masses, of the solute and of the solvent. Step three: compute the term the question asks for. And that's it! It works for every conversion, and if you forget a formula in the exam, you can rebuild it on the spot, just like we did today.

5:00Example 1: molality of 10% w/w NaOH

Right, let's try one. Find the molality of a ten percent, w by w, aqueous solution of sodium hydroxide. Which convenient amount will you take? Pause the video, and try it first. Done? Okay, let's check. It is given as a mass percent, so take one hundred grams of solution. Ten grams of it is sodium hydroxide. Its molar mass is forty, so ten by forty is zero point two five mole. Now the molality. Zero point two five mole in one hundred grams, so divide by zero point one kilogram. Two point five, right? Oh no, careful! Molality divides by the solvent, not the solution, remember? The water is one hundred minus ten, that is ninety grams, or zero point zero nine kilogram. So, small m equals zero point two five over zero point zero nine: two point seven eight molal. See, with the convenient amount, the whole question is just three lines.

6:04Example 2: molarity and molality of 98% H2SO4

Now the harder one, JEE pattern, and it is our puzzle from the start. Commercial sulphuric acid is ninety eight percent by mass, with density one point eight four grams per millilitre. Find its molarity and its molality. Pause the video and try. Hint: molarity is asked, so take one litre. Done? Okay, let's check. One litre is one thousand millilitres. Times one point eight four, that is one thousand eight hundred forty grams of solution. Ninety eight percent of it is acid: zero point nine eight times one thousand eight hundred forty, about one thousand eight hundred three grams. The molar mass of sulphuric acid is ninety eight, so that is eighteen point four moles. And they sit in one litre, so the molarity is eighteen point four molar. Let's quickly check with our first formula. Ten, times ninety eight, times one point eight four, over ninety eight. The ninety eights cancel, and we get eighteen point four again. Exactly the same! Now the molality. One litre of solution, so about one kilogram of water, so the molality is also eighteen point four, right? No, no! This is exactly the trap we saw. How much water is really there? One thousand eight hundred forty, minus one thousand eight hundred three: just thirty seven grams! That is zero point zero three seven kilogram. Almost all of the mass is acid. So, small m equals eighteen point four over zero point zero three seven. About five hundred moles per kilogram! And there it is, puzzle solved. Molality divides by the solvent only, and here there is hardly any solvent. So the molality shoots up to about five hundred, while the molarity stays at eighteen point four. Clear now?

8:01Quick revision: convenient amount, density, solvent only

So, quick revision. Pick a convenient amount: one litre of solution, one kilogram of solvent, or one hundred grams of solution. Density turns volume into mass, and molar mass turns grams into moles. And molality divides by the solvent only, never the whole solution. Now go and try the practice problems on this page. With this method, you will not need to mug up a single formula. All the best!