Class 11 Chemistry – Chapter 4: Stoichiometry (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Chapter 4 – Stoichiometry as per the Federal Board (FBISE) syllabus. Concepts are explained in a clear, student-friendly manner for better understanding and exam preparation.
Key topics covered include Mole and Molar Volume, Limiting and Non-Limiting Reactants, Theoretical Percentage Yield, detailed Stoichiometry problems, and solutions with step-by-step explanations for easy learning and revision.
Students can also access video lectures, MCQs, numericals, test series, and live classes for this chapter on our official YouTube channel and stay updated through our WhatsApp channel.
4.1 The Mole and Molar Volume
The mole is the standard unit used by chemists to count atoms, molecules, and ions. It provides a bridge between the microscopic world of particles and the macroscopic world of grams.
Key Concepts
- Avogadro's Number: One mole of any substance contains exactly $6.023 \times 10^{23}$ representative particles.
- Molar Mass: The mass of one mole of a substance in grams. This is numerically equal to the atomic mass, formula mass, or molecular weight.
- Example: 1 mole of $H_{2}O = 18\text{ g}$
- Example: 1 mole of $NaCl = 58.5\text{ g}$
- Molar Volume ($V_m$): One mole of any gas at Standard Temperature and Pressure (STP) occupies a volume of $22.414\text{ dm}^3$.
- Density of Gases: Since all gases occupy the same volume at STP ($22.414\text{ dm}^3$), the density of a gas is directly proportional to its molar mass.
Formulae Summary
$$\text{Volume} = \text{Moles} \times 22.414\text{ dm}^3$$ $$\text{Moles} = \frac{\text{Volume at STP}}{22.414\text{ dm}^3}$$ $$\rho = \frac{\text{Molar Mass}}{22.414\text{ dm}^3}$$| To Find | Formula/Relationship |
|---|---|
| Volume at STP | |
| Moles from Volume | |
| Density ($\rho$) at STP |
Concept Assessment Exercises
Q1. How many moles of oxygen molecules are there in $20.0\text{ dm}^3$ of oxygen gas at STP?
Solution:
We know that at STP, $22.414\text{ dm}^3$ of $O_2 = 1\text{ mole}$.
To find the moles in $20.0\text{ dm}^3$:
$$\text{Moles of } O_2 = \frac{20.0\text{ dm}^3}{22.414\text{ dm}^3/\text{mol}}$$ $$\text{Moles of } O_2 \approx 0.892\text{ moles}$$Q2. What volume does $0.6$ mole of $H_2$ gas occupy at STP?
Solution:
We know that $1\text{ mole}$ of $H_2$ at STP occupies $22.414\text{ dm}^3$.
To find the volume of $0.6$ moles:
$$\text{Volume} = 0.6\text{ moles} \times 22.414\text{ dm}^3/\text{mol}$$ $$\text{Volume} = 13.448\text{ dm}^3$$Q3. Calculate the gram molecular mass of a gas which has a density of $1.43\text{ g/dm}^3$ at STP.
Solution:
Density is mass per $1\text{ dm}^3$. At STP, molar mass is the mass of $22.4\text{ dm}^3$.
$$\text{Mass of } 1\text{ dm}^3 = 1.43\text{ g}$$ $$\text{Mass of } 22.4\text{ dm}^3 = 1.43\text{ g/dm}^3 \times 22.4\text{ dm}^3$$ $$\text{Molar Mass} = 32.032\text{ g/mol}$$The gram molecular mass is $32.032\text{ amu}$ (or grams per mole).
4.1.3 Stoichiometric Calculations and Mole Ratio
Stoichiometry involves using a balanced chemical equation to calculate the quantities of reactants and products. A balanced equation provides information on the macroscopic and microscopic levels.
Information from a Balanced Equation
Using the example: $2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)}$
- Molar Ratio: 2 moles of $H_2$ combine with 1 mole of $O_2$ to produce 2 moles of $H_2O$.
- Molecular Level: $2 \times 6.02 \times 10^{23}$ molecules of Hydrogen react with $6.02 \times 10^{23}$ molecules of Oxygen to produce $2 \times 6.02 \times 10^{23}$ molecules of water vapor.
- Mass Relationship: $4\text{ g}$ (2 moles) of $H_2$ combine with $32\text{ g}$ (1 mole) of $O_2$ to produce $36\text{ g}$ (2 moles) of $H_2O$.
- Volume at STP: $2 \times 22.414\text{ dm}^3$ of $H_2$ combine with $1 \times 22.414\text{ dm}^3$ of $O_2$ to produce $2 \times 22.414\text{ dm}^3$ of $H_2O$ vapor.
Dissociation of Ionic Compounds
When an ionic compound like $NaCl$ dissolves in water: $NaCl \xrightarrow{H_2O} Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
- 1 mole of $NaCl$ ($58.5\text{ g}$) gives 1 mole of $Na^{+}$ ions ($23\text{ g}$) and 1 mole of $Cl^{-}$ ions ($35.5\text{ g}$).
- $6.02 \times 10^{23}$ formula units of $NaCl$ produce $6.02 \times 10^{23}$ $Na^{+}$ ions and $6.02 \times 10^{23}$ $Cl^{-}$ ions.
Concept Assessment Exercise 4.2
Q1. What quantitative information do you get from the following chemical equation?
$$C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(g)}$$
Answer:
- Moles: 1 mole of Propane ($C_3H_8$) reacts with 5 moles of Oxygen to produce 3 moles of Carbon Dioxide and 4 moles of water vapor.
- Molecules: $6.02 \times 10^{23}$ molecules of $C_3H_8$ react with $5 \times 6.02 \times 10^{23}$ molecules of $O_2$.
- Volume (STP): $22.414\text{ dm}^3$ of $C_3H_8$ reacts with $112.07\text{ dm}^3$ ($5 \times 22.414$) of $O_2$.
Q2. Compare and contrast the terms: molecular mass and molar mass.
Answer:
- Molecular Mass: The sum of atomic masses of all atoms in a molecule, expressed in atomic mass units (amu).
- Molar Mass: The mass of one mole of a substance ($6.02 \times 10^{23}$ particles), expressed in grams/mole.
- Similarity: Both have the same numerical value for a given substance.
Q3. What mass of $Zn$ is needed to produce $100\text{ cm}^3$ of $H_2$ at STP?
$$Zn_{(s)} + 2HCl_{(aq)} \rightarrow ZnCl_{2(aq)} + H_{2(g)}$$
Solution:
- Convert Volume to $dm^3$: $100\text{ cm}^3 = 0.1\text{ dm}^3$.
- Find moles of $H_2$: $\frac{0.1\text{ dm}^3}{22.414\text{ dm}^3/\text{mol}} \approx 0.00446\text{ moles}$.
- Mole ratio from equation: $1\text{ mole } Zn : 1\text{ mole } H_2$.
- Mass of $Zn$: $0.00446\text{ moles} \times 65.38\text{ g/mol} \approx 0.29\text{ g}$.
4.1.3 Stoichiometric Calculations and Mole Ratio
A balanced chemical equation serves as a recipe for chemists, providing quantitative information on several levels. Using the mole as a central unit, we can calculate the mass, volume, and number of particles involved in a reaction.
Information from a Balanced Equation
Consider the reaction: $$2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)}$$
- Molar Ratio: 2 moles of $H_2$ react with 1 mole of $O_2$ to produce 2 moles of $H_2O$.
- Molecular Ratio: $2 \times 6.02 \times 10^{23}$ molecules of $H_2$ react with $6.02 \times 10^{23}$ molecules of $O_2$ to produce $2 \times 6.02 \times 10^{23}$ molecules of water vapor.
- Mass Ratio: $4\text{ g}$ of $H_2$ reacts with $32\text{ g}$ of $O_2$ to produce $36\text{ g}$ of $H_2O$.
- Volume Ratio (at STP): $2 \times 22.414\text{ dm}^3$ of $H_2$ reacts with $22.414\text{ dm}^3$ of $O_2$ to produce $2 \times 22.414\text{ dm}^3$ of $H_2O$ vapor.
Dissociation of Ionic Compounds
When ionic compounds dissolve in water, they separate into their constituent ions.
- Example ($NaCl$): 1 mole of $NaCl$ ($58.5\text{ g}$) yields 1 mole of $Na^{+}$ ($23\text{ g}$) and 1 mole of $Cl^{-}$ ($35.5\text{ g}$).
- Particle Count: $6.02 \times 10^{23}$ formula units of $NaCl$ produce $6.02 \times 10^{23}$ $Na^{+}$ ions and $6.02 \times 10^{23}$ $Cl^{-}$ ions.
4.1.4 Mole-Mole and Mole-Mass Calculations
Coefficients in a balanced equation indicate the proportion of moles between reactants and products.
- Mole-Mole: Determining the moles of one substance based on the known moles of another using the molar ratio.
- Mole-Mass: Converting the known mass of a reactant to moles, using the molar ratio to find the moles of the product, and then converting those moles back to mass.
Relevant Questions and Answers
Q1. What quantitative information is obtained from the equation: $C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(g)}$?
Answer:
- 1 mole of propane reacts with 5 moles of oxygen.
- The reaction produces 3 moles of carbon dioxide and 4 moles of water vapor.
- $22.414\text{ dm}^3$ of propane reacts with $112.07\text{ dm}^3$ ($5 \times 22.414$) of oxygen at STP.
Q2. How many moles of $CO$ are needed to reduce 5 moles of $Fe_2O_3$?
Equation: $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$
Solution:
- The molar ratio of $Fe_2O_3$ to $CO$ is $1:3$.
- For 5 moles of $Fe_2O_3$, the moles of $CO$ required = $5 \times 3 = 15\text{ moles}$.
Q3. Calculate the mass of $O_2$ produced if $25\text{ g}$ of $KClO_3$ is decomposed.
Equation: $$2KClO_3 \rightarrow 2KCl + 3O_2$$
Solution:
- Find Moles of $KClO_3$: $\text{Molar mass of } KClO_3 = 122.5\text{ g/mol}$. $\text{Moles} = \frac{25\text{ g}}{122.5\text{ g/mol}} = 0.204\text{ moles}$.
- Use Mole Ratio: 2 moles of $KClO_3$ give 3 moles of $O_2$. So, $0.204\text{ moles}$ give $\frac{3}{2} \times 0.204 = 0.306\text{ moles of } O_2$.
- Convert to Mass: $\text{Mass of } O_2 = 0.306\text{ mol} \times 32\text{ g/mol} = 9.796\text{ g}$.
Do You Know? Mole Day is celebrated on October 23rd from 6:02 am to 6:02 pm to commemorate Avogadro's constant ($6.02 \times 10^{23}$).
4.1.3 Stoichiometric Calculations and Mole Ratio
A balanced chemical equation provides quantitative relationships between reactants and products. Using the mole as a central unit, we can determine the mass, volume (at STP), and number of particles involved in a reaction.
Information from a Balanced Equation
Using the example: $$2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)}$$
- Molar Ratio: 2 moles of $H_2$ combine with 1 mole of $O_2$ to produce 2 moles of $H_2O$.
- Molecular Ratio: $2 \times 6.02 \times 10^{23}$ molecules of $H_2$ react with $6.02 \times 10^{23}$ molecules of $O_2$ to produce $2 \times 6.02 \times 10^{23}$ molecules of water vapor.
- Mass Relationship: $4\text{ g}$ (2 moles) of $H_2$ combine with $32\text{ g}$ (1 mole) of $O_2$ to produce $36\text{ g}$ (2 moles) of $H_2O$.
- Volume (at STP): $2 \times 22.414\text{ dm}^3$ of $H_2$ combine with $1 \times 22.414\text{ dm}^3$ of $O_2$ to produce $2 \times 22.414\text{ dm}^3$ of $H_2O$ vapor.
4.1.4 Mole-Mole and Mole-Mass Calculations
- Mole-Mole: Coefficients in a balanced equation indicate the proportion of moles of reactants and products. For example, in the burning of methane ($CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$), 1 mole of $CH_4$ requires 2 moles of $O_2$.
- Mole-Mass: If the mass of one substance is given, it is converted to moles, then the molar ratio is used to find the moles of the desired substance, which are then converted back to grams.
4.1.6 Solution Stoichiometry
In liquid solutions, concentration is commonly expressed as Molarity (M).
- Definition: Molarity is the number of moles of solute dissolved per $dm^3$ of solution.
- Formula: $$M = \frac{\text{moles of solute}}{\text{dm}^3 \text{ of solution}}$$
- Solution Stoichiometry: Volumes of solutions of known concentration are used instead of masses to determine the moles of reactants and products.
Relevant Questions and Answers
Q1. What quantitative information is obtained from the equation: $C_3H_{8(g)} + 5O_{2(g)} \rightarrow 3CO_{2(g)} + 4H_2O_{(g)}$?
Answer:
- 1 mole of $C_3H_8$ reacts with 5 moles of $O_2$.
- The reaction produces 3 moles of $CO_2$ and 4 moles of $H_2O$ vapor.
- Mass of 1 mole of propane ($44\text{ g}$) reacts with $160\text{ g}$ of oxygen ($5 \times 32\text{ g}$).
Q2. Compare and contrast molecular mass and molar mass.
Answer:
- Molecular mass is the mass of one molecule expressed in atomic mass units (amu).
- Molar mass is the mass of one mole ($6.02 \times 10^{23}$ particles) of a substance expressed in grams/mole.
- Both values are numerically identical for a specific substance.
Q3. How many moles of $CO$ are needed to reduce 5 moles of $Fe_2O_3$?
Equation: $Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$
Solution:
- According to the equation, 1 mole of $Fe_2O_3$ requires 3 moles of $CO$.
- Therefore, 5 moles of $Fe_2O_3$ require: $5 \times 3 = 15\text{ moles of } CO$.
Q4. Calculate the molarity of a solution containing $40\text{ g}$ of urea ($NH_2CONH_2$) in $500\text{ cm}^3$ of solution.
Solution:
- Molar mass of urea: $14 + 1 \times 2 + 12 + 16 + 14 + 1 \times 2 = 60\text{ g/mol}$.
- Moles of urea: $\frac{40\text{ g}}{60\text{ g/mol}} = 0.667\text{ moles}$.
- Volume in $dm^3$: $\frac{500\text{ cm}^3}{1000} = 0.5\text{ dm}^3$.
- Molarity: $\frac{0.667\text{ mol}}{0.5\text{ dm}^3} = 1.334\text{ M}$.
4.1.3 Stoichiometric Calculations and Mole Ratio
Stoichiometry allows us to calculate the mass, volume, and number of particles of reactants and products based on a balanced chemical equation.
Information from a Balanced Equation
A balanced equation like $$2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)}$$ provides the following:
- Moles: 2 moles of $H_2$ react with 1 mole of $O_2$ to produce 2 moles of $H_2O$.
- Molecules: $2 \times 6.02 \times 10^{23}$ molecules of $H_2$ react with $6.02 \times 10^{23}$ molecules of $O_2$.
- Mass: $4\text{ g}$ of $H_2$ reacts with $32\text{ g}$ of $O_2$ to produce $36\text{ g}$ of $H_2O$.
- Volume at STP: $2 \times 22.414\text{ dm}^3$ of $H_2$ reacts with $22.414\text{ dm}^3$ of $O_2$.
4.1.4 Mole-Mole and 4.1.5 Mole-Mass Calculations
- Mole-Mole: Coefficients in a balanced equation indicate the molar proportions. For example, in $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$, 1 mole of $CH_4$ requires 2 moles of $O_2$.
- Mole-Mass: If the mass of a substance is given, convert it to moles, use the molar ratio to find the moles of the product, and then convert back to mass.
4.1.6 Solution Stoichiometry
In solutions, concentration is measured in Molarity (M), which is the number of moles of solute per $dm^3$ of solution.
$$\text{Molarity (M)} = \frac{\text{mole of solute}}{\text{dm}^3 \text{ of solution}}$$
For reactions between two solutions, the following formula is used: $$\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$$
- $M_1, M_2$: Molarities of reactants 1 and 2.
- $V_1, V_2$: Volumes of reactants 1 and 2.
- $n_1, n_2$: Moles of reactants from the balanced equation.
Relevant Questions and Answers
Q1. How many moles of $CO$ are needed to reduce 5 moles of $Fe_2O_3$?
Equation: $$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$$
Answer: According to the balanced equation, 1 mole of $Fe_2O_3$ requires 3 moles of $CO$. Therefore, 5 moles of $Fe_2O_3$ will require $5 \times 3 = 15 \text{ moles of } CO$.
Q2. Calculate the mass of $O_2$ produced by the decomposition of $25\text{ g}$ of $KClO_3$.
Answer:
- Molar mass of $KClO_3 = 122.5\text{ g/mol}$.
- Moles of $KClO_3 = \frac{25}{122.5} = 0.204 \text{ moles}$.
- From the equation $2KClO_3 \rightarrow 2KCl + 3O_2$, 2 moles of $KClO_3$ give 3 moles of $O_2$.
- Moles of $O_2 = \frac{3}{2} \times 0.204 = 0.306 \text{ moles}$.
- Mass of $O_2 = 0.306 \times 32 = 9.796\text{ g}$.
Q3. What volume of $0.5\text{ M } Na_2SO_4$ reacts with $275\text{ cm}^3$ of $0.25\text{ M } BaCl_2$?
Equation: $$Na_2SO_4 + BaCl_2 \rightarrow BaSO_4 + 2NaCl$$
Answer:
- $M_1 = 0.5\text{ M}, n_1 = 1$.
- $M_2 = 0.25\text{ M}, V_2 = 275\text{ cm}^3, n_2 = 1$.
- Using $\frac{0.5 \times V_1}{1} = \frac{0.25 \times 275}{1}$.
- $V_1 = \frac{0.25 \times 275}{0.5} = 137.5\text{ cm}^3$.
Do You Know? Mole Day is celebrated on October 23 between 6:02 am and 6:02 pm to represent Avogadro's constant, $6.02 \times 10^{23}$.
4.2 Limiting and Non-Limiting Reactants
When two reactants are mixed for a reaction, usually one is completely consumed while the other is left over.
- Limiting Reactant: The reactant that is completely consumed and limits the amount of product formed.
- Non-Limiting (Excess) Reactant: The reactant that remains unreacted after the completion of the reaction.
Example: Hydrogen and Oxygen Reaction
Consider the reaction: $$2H_2 + O_2 \rightarrow 2H_2O$$
If 1 mole of $H_2$ and 1 mole of $O_2$ are mixed:
- 1 mole of $H_2$ requires only $0.5$ moles of $O_2$.
- Since 1 mole of $O_2$ is available, it is in excess.
- $H_2$ is the limiting reactant because it is consumed first.
Stoichiometry Calculation Summary
Using the mole as a central unit, several quantitative relationships can be established:
| Conversion Type | Key Relationship/Formula |
|---|---|
| Mole to Particles | $1\text{ mole} = 6.023 \times 10^{23}$ representative particles (Avogadro's Number) |
| Mole to Volume (STP) | $1\text{ mole of any gas} = 22.414\text{ dm}^3$ |
| Mole to Mass | $\text{Mass} = \text{Moles} \times \text{Molar Mass}$ |
| Solution Molarity | $M = \frac{\text{mole of solute}}{\text{dm}^3 \text{ of solution}}$ |
Relevant Questions and Answers
Q1. How much volume is occupied by marsh gas ($CH_4$) at STP containing $4.8 \times 10^{24}$ molecules?
Answer:
- Find Moles: $\text{Moles} = \frac{4.8 \times 10^{24}}{6.023 \times 10^{23}} \approx 7.97\text{ moles}$.
- Find Volume: $\text{Volume} = 7.97\text{ moles} \times 22.414\text{ dm}^3/\text{mol} \approx 178.6\text{ dm}^3$.
Q2. Calculate the mass of calcium nitride ($Ca_3N_2$) prepared from $54.9\text{ g}$ of $Ca$ and $43.2\text{ g}$ of $N_2$.
Equation: $3Ca + N_2 \rightarrow Ca_3N_2$
Solution:
- Moles of $Ca$: $\frac{54.9}{40} = 1.37\text{ moles}$.
- Moles of $N_2$: $\frac{43.2}{28} = 1.54\text{ moles}$.
- Identify Limiting Reactant: 3 moles of $Ca$ produce 1 mole of $Ca_3N_2$. Thus, $1.37$ moles of $Ca$ produce $\frac{1.37}{3} = 0.457\text{ moles of } Ca_3N_2$.
- $N_2$ would produce more, so $Ca$ is the limiting reactant.
- Mass of $Ca_3N_2$: $0.457\text{ moles} \times 148\text{ g/mol} \approx 67.6\text{ g}$.
Q3. What is Mole Day?
Answer: Mole Day is a celebration for chemists held on October 23 (10/23) between 6:02 AM and 6:02 PM to commemorate Avogadro's constant ($6.02 \times 10^{23}$).
Q4. Calculate the molarity of a solution containing $75\text{ g}$ of $KClO_3$ in $1.25\text{ dm}^3$ of solution.
Answer:
- Molar mass of $KClO_3$: $122.5\text{ g/mol}$.
- Moles of $KClO_3$: $\frac{75}{122.5} \approx 0.612\text{ moles}$.
- Molarity: $\frac{0.612\text{ mol}}{1.25\text{ dm}^3} \approx 0.49\text{ M}$.
4.3 Theoretical Yield, Actual Yield, and Percent Yield
In chemical reactions, the amount of product we expect to get based on calculations is often different from what we actually obtain in the laboratory.
- Theoretical Yield: The maximum amount of product that can be produced from a given amount of reactant according to a balanced chemical equation. It assumes a 100% conversion of the limiting reactant.
- Actual Yield: The amount of product actually produced experimentally in a chemical reaction. This is usually less than the theoretical yield due to side reactions, products remaining in solution, or reactions stopping before completion.
- Percent Yield: The ratio of the actual yield to the theoretical yield, expressed as a percentage.
Key Formula
To calculate the efficiency of a reaction, we use the following formula:
$$\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$Concept Assessment Exercise 4.6
Q1. Calculate the percent yield of Ozone ($O_3$) if $10\text{ g}$ of $O_2$ produces $1.05\text{ g}$ of $O_3$.
Equation: $3O_2 \rightarrow 2O_3$
Solution:
- Find Moles of $O_2$: $\text{Molar mass of } O_2 = 32\text{ g/mol}$. $\text{Moles} = \frac{10\text{ g}}{32\text{ g/mol}} = 0.3125\text{ moles}$.
- Use Mole Ratio: From the equation, $3\text{ moles of } O_2$ give $2\text{ moles of } O_3$. Therefore, $0.3125\text{ moles of } O_2$ will give $\frac{2}{3} \times 0.3125 = 0.208\text{ moles of } O_3$.
- Find Theoretical Yield: $\text{Molar mass of } O_3 = 48\text{ g/mol}$. $\text{Mass} = 0.208\text{ moles} \times 48\text{ g/mol} = 10\text{ g}$.
- Calculate Percent Yield: $\frac{1.05\text{ g (Actual)}}{10\text{ g (Theoretical)}} \times 100 = 10.5\%$.
Q2. Baking soda ($NaHCO_3$) is prepared by passing $NH_3$ and $CO_2$ through $NaCl$ solution. If $20\text{ g } NH_3$ and $30\text{ g } CO_2$ produce $40\text{ g}$ baking soda, calculate the percent yield.
Equation: $NaCl + H_2O + CO_2 + NH_3 \rightarrow NaHCO_3 + NH_4Cl$
Answer Note: To solve this, you must first identify the limiting reactant between $NH_3$ and $CO_2$, calculate the theoretical yield of $NaHCO_3$, and then use the percent yield formula with the $40\text{ g}$ actual yield.
Summary Table: Stoichiometry Units
| Quantity | Unit/Value at STP |
|---|---|
| 1 Mole (Particles) | $6.023 \times 10^{23}$ |
| Molar Volume ($V_m$) | $22.414\text{ dm}^3$ |
| Density Calculation | $\frac{\text{Molar Mass}}{22.414\text{ dm}^3}$ |