Advance Concepts and Questions of MOLE | Zero Chemistry Course | Class 11, Chapter 1 | Lecture 3

Ajouté : 2026-05-05

[00:00:00]In the name of Allah, the most beneficent, the most merciful. We will solve some advanced questions of mole, including competitive exam questions. To learn and solve these advanced questions, I will teach you my favorite method for mole calculation, which will clear your all confusion.

[00:00:21]For example, consider 10 pencils. Let the price of 10 pencils is 20 rupees. If I ask you, what is the price of five pencils? You can instantly say 10 rupees. But wait a minute.

[00:00:37]Let me show you how we use mathematical method to calculate it. Here, listen carefully. I say the price of 10 pencils is 20 rupees. Then the price of five pencils is unknown or X. I mean, we know the price of 10 pencils, but the price of five pencils is unknown. For unknown value, we use X. Now, to find the value of X, I cross multiply it. 10 into X equals to five into 20. I divide both sides by 10. After calculation, I get X is equal to 10 rupees. Thus, the price of five pencils is 10 rupees.

[00:01:25]Now, we will use this method to calculate moles in a chemical reaction.

[00:01:30]For example, consider this reaction.

[00:01:34]Here, this two is known as stoichiometric coefficient of hydrogen gas. This one is known as stoichiometric coefficient of oxygen gas. And this two is known as stoichiometric coefficient of water. Remember that stoichiometric coefficient tells us mole ratio between reactants and products. Let me repeat it. Stoichiometric coefficient tells us mole ratio between reactants and products. Here, there are three substances, like hydrogen gas, oxygen gas, and water. The mole ratio between them is 2:1:2.

[00:02:16]I mean, stoichiometric coefficient tells us that two moles of hydrogen gas react with one mole of oxygen gas to form two moles of water.

[00:02:28]Now, to master this concept, consider this question. How many moles of oxygen is required to react with four moles of hydrogen gas? Let me show you the easy way to calculate such type of numerical problems.

[00:02:44]First of all, always spot the substances in the numerical problem, like oxygen gas and hydrogen gas. So, we will create stoichiometric ratio between oxygen gas and hydrogen gas. According to the reaction, we can see that the stoichiometric coefficient of hydrogen gas is two and that of oxygen gas is one. So, the mole ratio between them is 2:1.

[00:03:13]Secondly, I search for mole values in the question. We can see that four moles of hydrogen gas is given. So, I write four moles of hydrogen gas. Now, I say according to the stoichiometric values, two moles of hydrogen gas react with one mole of oxygen gas, then four moles of hydrogen gas react with X mole of oxygen gas. To find the value of X, I cross multiply them. I get [clears throat] 2X equals to four into one. After calculation, I get two moles of oxygen gas. It means that two moles of hydrogen gas react with one mole of oxygen gas, while four moles of hydrogen gas react with two moles of oxygen gas. Hence, note it down this very, very important method of calculation, which I will use in complete stoichiometry.

[00:04:17]Now, let me show you what I mean by advanced question. For example, consider this question. In the question, we are asked to find the number of molecules present in 10 g of common salt that contains 96% sodium chloride by mass.

[00:04:37]Well, there are three things in the question. We have the percentage of sodium chloride, we have the total mass, and we need to find the number of molecules.

[00:04:48]So, we will solve this question in three steps.

[00:04:51]In the first step, we will find the actual mass of sodium chloride present in 10 g of salt. This 96% sodium chloride means in 100% sea water, there are 96% sodium chloride present.

[00:05:10]Here, the given mass of sea water is 10 g. So, we will calculate mass of sodium chloride in 10 g of sea water.

[00:05:20]I write in 100% sea water, there are 96% sodium chloride present. The given mass of sea water is 10 g. So, I say in 10 g of sea water, X g of sodium chloride is present.

[00:05:38]To calculate value of X, I cross multiply them. I get 100 multiplied by X and 96 by 10. After calculation, I get 9.6 g.

[00:05:53]It means that in 10 g of sea water, there are 9.6 g of sodium chloride present.

[00:06:01]Now, in the second step, we will convert grams of sodium chloride to moles of sodium chloride.

[00:06:08]We can see that key number two has mass and moles. So, we will use key number two. Mole is equal to given mass upon gram molecular mass. The gram molecular mass of sodium chloride is 23 + 35.5, which equals 58.5 g per mole.

[00:06:29]So, I write 9.6 upon 58.5.

[00:06:34]After calculation, I get 0.164 moles of sodium chloride.

[00:06:42]Now, in the third step, we will convert moles to number of molecules. We can see that key number three has moles and molecules.

[00:06:52]So, we will use key number three. I write 0.164 multiplied by 6.022 * 10 raised to the power 23. After calculation, I get 0.987 * 10 raised to the power 23. And 0.987 * 10 raised to the power 23 is approximately equal to 10 raised to the power 23. Thus, the correct option is C.

[00:07:24]Hence, note it down this question.

[00:07:28]Now, consider this NEET question of 2019.

[00:07:32]In this question, we are asked to find the number of moles of hydrogen required to produce 20 moles of ammonia through Haber's process.

[00:07:42]Well, there are two things in this question. Moles of ammonia are given, and we need to find moles of hydrogen.

[00:07:50]Now, pay attention here. We don't need any key here. Just simple calculation based on the reaction.

[00:07:59]In the first step, we will write the balanced equation for Haber's process.

[00:08:04]Nitrogen gas plus three hydrogen gas gives two moles of ammonia.

[00:08:10]Now, look at the stoichiometric coefficients. The stoichiometric coefficient of hydrogen gas is three, and the stoichiometric coefficient of ammonia is two. By stoichiometric coefficient, I mean the number written in front of each molecule in the balanced equation.

[00:08:28]Now, in the question, there is hydrogen gas and there is ammonia. So, I will create ratio between hydrogen gas and ammonia. According to the reaction, the stoichiometric coefficient of hydrogen gas is three and that of ammonia is two.

[00:08:45]So, the ratio between hydrogen gas to ammonia is 3:2.

[00:08:51]In the second step, we will do the simple calculation. According to the reaction, two moles of ammonia requires three moles of hydrogen gas. So, 20 moles of ammonia will require X mole.

[00:09:06]I cross multiply them. I write X is equal to 3 upon two multiplied by 20.

[00:09:14]After calculation, I get X is equal to 30 moles of hydrogen gas. It means that 20 moles of ammonia require 30 moles of hydrogen gas. Thus, the correct option is option C, 30 moles. Hence, note it down.

[00:09:35]Now, consider this question.

[00:09:38]If you learn this question, you will be able to solve two competitive exam questions, one from NEET and one from JEE Mains. In this question, we are asked to find the ratio of the number of molecules of carbon dioxide gas and nitrous oxide in a gaseous mixture where their masses are in the ratio 2 to 5.

[00:10:04]Well, there are two things in this question. Mass ratio is given and we need to find the molecule ratio.

[00:10:12]So, we will solve this question in two steps. Now, pay attention here. Before we start, there is a golden rule for all ratio questions.

[00:10:22]The mole ratio of substances is equal to the molecule ratio of substances.

[00:10:28]Let me repeat it. Mole ratio equals molecule ratio. So, we do not need to find actual molecules.

[00:10:38]We just need to find the mole ratio and that is enough.

[00:10:42]In the second step, we will write the molecular masses. The gram molecular masses of carbon dioxide gas is 44 g per mole. The gram molecular mass of nitrous oxide is 44 g per mole. Notice that both have the same molecular masses.

[00:11:02]In the second step, we will find the moles of each gas using key number two.

[00:11:07]Mole is equal to given mass upon gram molecular mass. The given mass of carbon dioxide is two and the given mass of nitrous oxide is five. These are ratio values, not actual grams, but that is perfectly fine. So, I write moles of carbon dioxide equals to 2 upon 44 and moles of nitrous oxide equals to 5 upon 44.

[00:11:37]We can see that at both sides, 44 is common. So, this 44 cancels from both side. After calculation, I get the ratio is 2 to 5.

[00:11:51]It means the mole ratio between carbon dioxide gas and nitrous oxide is 2 to 5.

[00:11:58]We already learned that mole ratio equals molecule ratio.

[00:12:04]The ratio of molecules of carbon dioxide to nitrous oxide is also 2 to 5. Thus, the correct answer is option B. Hence, note it down this important question.

[00:12:20]Now, we will solve exactly the same type of question from NEET. For example, consider this question. Pause the video and solve it. Don't worry if you make mistakes. I am here to correct them.

[00:12:36]Well, there are two things. Mass ratio is given and we need to find the molar ratio. The ratio of masses of hydrogen gas to oxygen gas is 1 to 4. So, we just need to find the mole ratio. That's enough.

[00:12:53]In the first step, we will write the molecular masses.

[00:12:57]Now, pay attention here. Hydrogen and oxygen are both diatomic gases.

[00:13:03]By diatomic, I mean they exist as H2 and O2, not as single atoms.

[00:13:11]The gram molecular mass of hydrogen gas is two. The gram molecular mass of oxygen gas is 32 g. In the second step, we will find the moles of each gas using key number two. Mole is equal to given mass upon gram molecular mass. The given mass of hydrogen gas is one and the given mass of oxygen gas is four. These are ratio values, not actual grams, but that is perfectly fine. So, I write moles of hydrogen gas equals 1 upon 2 and moles of oxygen gas equals 4 upon 32.

[00:13:51]Now, listen carefully. This ratio sign means division. Let me repeat it. This ratio sign means division. I write 1 upon 2. Then, I remove the ratio symbol and I put a division symbol.

[00:14:09]Here, we change the denominator to numerator. To do so, we change the division symbol to multiplication symbol.

[00:14:18]I write 1 upon 2 multiplied by 32 upon 4. After calculation, I get 32 upon 8.

[00:14:29]Now, I simplify this fraction. I write 4 upon 1.

[00:14:35]I can also write it as 4 ratio 1.

[00:14:39]I mean, mole ratio of hydrogen gas to oxygen gas is 4 to 1.

[00:14:45]Thus, the correct option is A. Hence, note it down this NEET question.

[00:14:53]Now, we will solve exactly the same type of questions from JEE Main exam.

[00:14:59]Consider this question from 2014.

[00:15:03]Pause the video and try to solve it.

[00:15:06]Don't worry if you make mistakes. I am here to correct them.

[00:15:12]Well, in this question, we are asked to find the ratio of the number of molecules of oxygen and nitrogen in a gaseous mixture. The ratio of their masses are 1 to 4.

[00:15:25]Here, students do blunders. Students take atoms of oxygen and nitrogen, which are totally wrong. We are asked to find the number of molecules. So, we will take gas of oxygen and gas of nitrogen.

[00:15:41]Now, there are two things in the question. Mass ratio is given and we need to find the molecule ratio.

[00:15:48]So, we will solve this question in two steps. In the first step, we will write the molecular masses.

[00:15:55]Now, the gram molecular mass of oxygen gas is 32 g. The gram molecular mass of nitrogen gas is 28 g.

[00:16:05]In the second step, we will find the moles of each gas using key number two.

[00:16:10]Mole is equal to given mass upon gram molecular mass. The given mass of oxygen gas is one and the given mass of nitrogen gas is four. These are ratio values, not actual grams, but that is perfectly fine. So, I write moles of oxygen gas equals 1 upon 32. And moles of nitrogen gas equals 4 upon 28.

[00:16:38]Now, this ratio sign means division.

[00:16:41]I write 1 upon 32. Then, I remove the ratio symbol and I put a division symbol. By changing the denominator to numerator, we can also change the division symbol to multiplication symbol.

[00:16:57]I write 1 upon 32 multiplied by 28 upon 4.

[00:17:03]After calculation, I get 28 upon 128.

[00:17:10]I divide both by 4. I reduce it to 7 upon 32.

[00:17:15]I can also write it as 7 ratio 32. Thus, the ratio of moles between oxygen gas and nitrogen gas is 7 to 32. We already learned that mole ratio equals molecule ratio. So, the ratio of molecules of oxygen gas to nitrogen gas is also 7 to 32.

[00:17:39]Thus, the correct option is B.

[00:17:42]Hence, note it down this JEE question.

[00:17:47]Now, consider this another JEE question from 2019 and try to solve it.

[00:17:54]Well, in this question, we are asked to find the mass of water produced from 445 g of this substance. There are two things in the question. Mass of this substance is given and we need to find the mass of water. We will solve this question in four steps. In the first step, we will find the molecular masses.

[00:18:19]The molecular mass of this compound is 890 g per mole.

[00:18:25]The gram molecular mass of water is 18 g.

[00:18:29]In the second step, we will find the number of moles of this substance using key number two. Mole is equal to given mass upon gram molecular mass. So, I write the given mass of this molecule is 445 g upon the gram molecular mass of this compound is 890.

[00:18:52]After calculation, I get 0.5 moles of this compound.

[00:18:59]Now, in the third step, we will use the balanced equation. The balanced equation tells us that two moles of this compound gives 110 moles of water. Let me repeat it. Two moles of this compound gave 110 moles of water. So, if two mole gives 110 moles, then 0.5 moles will give X moles.

[00:19:23]I write X equals to 110 multiplied by 0.5 upon 2. After calculation, I get 27.5 moles of water. Thus, we get 27.5 moles of water.

[00:19:42]In the fourth step, we will find the mass of water using key number two.

[00:19:48]Key number two normally says mole equals given mass upon gram molecular mass. I rearrange it. Mass equals moles multiplied by gram molecular mass.

[00:20:01]I write 27.5 moles multiplied by gram molecular mass, which is 18 g. After calculation, I get 495 g of water.

[00:20:17]Thus, the correct option is B, 495 g. Hence, note it down this important question.

[00:20:26]In the next lecture, we will learn all about limiting reagent.