Class 11 Chap 01 : Some Basic Concept Of Chemistry 03 : MOLARITY and MOLALITY || MOLARITY|| MOLALITY

TLDR;

This video by Physics Wallah - Alakh Pandey provides a comprehensive explanation of molarity and molality, two important concepts in chemistry. It begins with definitions and formulas, then walks through several example problems of increasing complexity, and concludes with a discussion of molarity of dilution and molarity of mixture.

Molarity is defined as the number of moles of solute per liter of solution.
Molality is defined as the number of moles of solute per kilogram of solvent.
The video includes example problems that cover a range of scenarios, including calculating molarity from mass and volume, calculating mass from molarity and volume, and calculating molarity of ions in solution.

Introduction [0:00]

The video introduces the concepts of molarity and molality, addressing common student concerns about molarity. The instructor assures viewers that the topic is straightforward and aims to clarify it within 30 minutes.

Understanding Molarity [0:27]

Molarity is defined as the number of moles of solute present in one liter of solution. For example, a 2M sugar solution contains two moles of sugar in one liter of solution. Similarly, a 3M solution of H+ ions contains three moles of H+ ions in one liter of solution. The formula for molarity is: Molarity = Number of moles of solute / Volume of solution (in liters). The number of moles can be calculated as: Number of moles = Given mass / Molecular mass. The unit for molarity is moles per liter, denoted as M.

Example Problem 1: Calculating Molarity of NaOH Solution [3:45]

The video demonstrates how to calculate the molarity of a solution containing 40g of NaOH dissolved in 250 ml of solution. The atomic masses of Na, O, and H are given as 23, 16, and 1, respectively. The molarity is calculated using the formula: Molarity = (Mass of solute / Molecular mass of solute) / Volume of solution in liters. The number of moles of NaOH is calculated as 40g / 40g/mol = 1 mol. The volume of the solution is converted to liters: 250 ml = 0.25 liters. Therefore, the molarity of the solution is 1 mol / 0.25 liters = 4 M.

Example Problem 2: Calculating Molarity of Sulfuric Acid Solution [5:23]

The video explains how to find the molarity of 4.9 grams of sulfuric acid (H2SO4) present in 500 cubic centimeters of solution. The molarity formula is Molarity = (mass / molecular mass) / volume of solution in liters. The mass is 4.9 grams. The molecular mass of H2SO4 is calculated as (1 * 2) + 32 + (16 * 4) = 98 g/mol. The volume is 500 cm³, which is converted to liters by dividing by 1000, resulting in 0.5 liters. Molarity = (4.9 / 98) / 0.5 = 0.1 M.

Example Problem 3: Finding Mass of Sodium Carbonate in a Solution [7:15]

The video demonstrates how to calculate the mass of Na2CO3 present in 100 ml of a 3M solution. The molarity formula is used: Molarity = Number of moles / Volume of solution in liters. Given the molarity (3M) and volume (100 ml), the number of moles of Na2CO3 is calculated. First, convert the volume to liters: 100 ml = 0.1 liters. Then, 3 = Number of moles / 0.1, so Number of moles = 0.3. Next, the mass is calculated using the formula: Mass = Number of moles * Molecular mass. The molecular mass of Na2CO3 is (23 * 2) + 12 + (16 * 3) = 106 g/mol. Therefore, Mass = 0.3 * 106 = 31.8 grams.

Example Problem 4: Molarity with Weight by Weight Percentage [10:18]

The video explains how to calculate the molarity of a 10% weight by weight aqueous solution of H2SO4, given that the density of the solution is 1.1 g/ml. A 10% weight by weight solution means 10 grams of solute (H2SO4) are present in 100 grams of solution. The molarity formula is Molarity = (Number of moles of solute) / (Volume of solution in liters). The number of moles of H2SO4 is calculated as 10 g / 98 g/mol (molecular mass of H2SO4). To find the volume of the solution, the density formula is used: Density = Mass / Volume. Volume = Mass / Density = 100 g / 1.1 g/ml = 90.91 ml, which is then converted to liters. Finally, the molarity is calculated using the values obtained.

Example Problem 5: Molarity with Solute and Solvent Masses [13:46]

The video explains how to calculate the molarity of a solution containing 120 grams of urea in 1000 grams of water, given that the density of the solution is 1.12 g/ml. The molarity formula is Molarity = (Number of moles of solute) / (Volume of solution in liters). The number of moles of urea is calculated as 120 g / 60 g/mol (molecular mass of urea) = 2 moles. To find the volume of the solution, the mass of the solution is calculated by adding the mass of the solute and the solvent: 120 g + 1000 g = 1120 g. Then, the density formula is used: Density = Mass / Volume. Volume = Mass / Density = 1120 g / 1.12 g/ml = 1000 ml = 1 liter. Therefore, the molarity of the solution is 2 moles / 1 liter = 2 M.

Special Case: Molarity of Water [16:41]

The video addresses a common question about the molarity of water. It emphasizes that the molarity of water is approximately 55.55 M only when the density of water is normal (1 g/ml or 1 kg/L). The calculation involves considering 100 ml of water. Since density is 1 g/ml, the mass of water is 100 g. The number of moles is calculated as 100 g / 18 g/mol (molecular mass of water). The volume is 100 ml, which is converted to 0.1 liters. Molarity = (100/18) / 0.1 = 55.55 M. If the density is different, such as 0.9 g/ml, the mass would be 90 g, and the molarity would need to be recalculated.

Molar, Semimolar, Decimolar, and Centimolar Solutions [20:07]

The video defines different types of solutions based on their molarity:

Molar solution: Molarity is 1 (1 M).
Semimolar solution: Molarity is 1/2 (0.5 M or M/2).
Decimolar solution: Molarity is 1/10 (0.1 M or M/10).
Centimolar solution: Molarity is 1/100 (0.01 M or M/100).

Example Problem 6: Mass of KOH in a Semimolar Solution [21:39]

The video explains how to find the mass of potassium hydroxide (KOH) present in 150 cm³ of a semimolar aqueous solution. Semimolar means the molarity is 1/2 M. The molarity formula is Molarity = Number of moles / Volume of solution in liters. Given the molarity (0.5 M) and volume (150 cm³), the number of moles of KOH is calculated. First, convert the volume to liters: 150 cm³ = 0.15 liters. Then, 0.5 = Number of moles / 0.15, so Number of moles = 0.075. Next, the mass is calculated using the formula: Mass = Number of moles * Molecular mass. The molecular mass of KOH is approximately 36 g/mol. Therefore, Mass = 0.075 * 36 = 2.7 grams.

Molarity of Ions [23:59]

The video introduces the concept of molarity of ions, focusing on how to calculate the molarity of individual ions in a solution.

Example Problem 7: Molarity of Ions in Sodium Carbonate Solution [24:10]

The video explains how to find the molarity of sodium ions (Na+) and carbonate ions (CO3^2-) in a solution containing 53 grams of sodium carbonate (Na2CO3) dissolved in 500 ml of solution. First, the molarity of Na2CO3 is calculated: Molarity = (Number of moles of solute) / (Volume of solution in liters). The number of moles of Na2CO3 is calculated as 53 g / 106 g/mol (molecular mass of Na2CO3) = 0.5 moles. The volume is 500 ml = 0.5 liters. Therefore, the molarity of Na2CO3 is 0.5 moles / 0.5 liters = 1 M. Next, the dissociation of Na2CO3 is considered: Na2CO3 dissociates into 2 Na+ ions and 1 CO3^2- ion. Therefore, for every 1 mole of Na2CO3, there are 2 moles of Na+ ions and 1 mole of CO3^2- ions. The molarity of Na+ ions is 2 M, and the molarity of CO3^2- ions is 1 M.

Example Problem 8: Molarity of Hydrogen Ions in Sulfuric Acid Solution [28:09]

The video presents a problem to calculate the molarity of H+ ions if 49 grams of H2SO4 is dissolved in 250 ml solution. H2SO4 dissociates into 2H+ and SO4^2-. The molarity of H2SO4 is calculated first, and then the molarity of H+ ions is twice the molarity of H2SO4 because each molecule of H2SO4 produces two H+ ions.

Molarity of Dilution [28:59]

The video introduces the concept of molarity of dilution, explaining how the molarity of a solution changes when water is added. Adding water increases the volume of the solution, which decreases the molarity. The key principle is that the number of moles of solute remains constant during dilution. The formula for dilution is M1V1 = M2V2, where M1 and V1 are the initial molarity and volume, and M2 and V2 are the final molarity and volume after dilution.

Derivation of Dilution Formula [30:30]

The video explains the derivation of the dilution formula (M1V1 = M2V2). The number of moles of solute remains constant during dilution. Molarity is defined as Number of moles of solute / Volume of solution. Therefore, Number of moles of solute = Molarity * Volume. Since the number of moles is constant, M1V1 = M2V2.

Example Problem 9: Calculating New Molarity After Dilution [32:29]

The video explains how to calculate the new molarity when 0.2 molar 100 ml HNO3 solution is diluted with 100 ml water. The initial molarity (M1) is 0.2 M, and the initial volume (V1) is 100 ml. After adding 100 ml of water, the final volume (V2) is 200 ml. Using the dilution formula M1V1 = M2V2, the final molarity (M2) is calculated: 0.2 * 100 = M2 * 200. Solving for M2, M2 = (0.2 * 100) / 200 = 0.1 M.

Molarity of Mixture [33:52]

The video introduces the concept of molarity of mixture, explaining how to calculate the molarity when two solutions containing the same solute are mixed. The formula for molarity of mixture is M = (M1V1 + M2V2) / (V1 + V2), where M1 and V1 are the molarity and volume of the first solution, and M2 and V2 are the molarity and volume of the second solution.

Derivation of Mixture Formula [35:33]

The video explains the derivation of the mixture formula. The total number of moles in the mixture is the sum of the number of moles in each solution. Number of moles = Molarity * Volume. Therefore, M1V1 + M2V2 = M(V1 + V2). Solving for M, M = (M1V1 + M2V2) / (V1 + V2).

Example Problem 10: Calculating Molarity of Mixture [37:09]

The video explains how to find the molarity of a mixture when 0.2 molar 100 ml HCl solution is mixed with 0.1 molar 300 ml HCl solution. Using the formula M = (M1V1 + M2V2) / (V1 + V2), the molarity of the mixture is calculated: M = (0.2 * 100 + 0.1 * 300) / (100 + 300) = (20 + 30) / 400 = 50 / 400 = 0.125 M.

Understanding Molality [38:25]

The video introduces the concept of molality, distinguishing it from molarity. Molality is defined as the number of moles of solute present in 1 kg of solvent, whereas molarity is the number of moles of solute present in 1 liter of solution. The formula for molality is: Molality = Number of moles of solute / Mass of solvent (in kg).

Molality Formula and Units [39:31]

The video reiterates the molality formula: molality = number of moles of solute / mass of solvent (in kg). It emphasizes that molality involves the mass of the solvent, not the volume of the solution. The units for molality are moles per kilogram, denoted as m.

Example: Molality of HCl Solution [40:34]

The video provides an example: a 3m solution of HCl means there are 3 moles of HCl present in 1 kg of water (solvent).

Example Problem 11: Calculating Molality of NaOH Solution [41:20]

The video explains how to calculate the molality of a solution containing 20 g of NaOH in 100 g of solution. First, the number of moles of NaOH is calculated: 20 g / 40 g/mol (molecular mass of NaOH) = 0.5 moles. Next, the mass of the solvent (water) is calculated: 100 g (solution) - 20 g (NaOH) = 80 g. This is converted to kg: 80 g = 0.08 kg. Finally, the molality is calculated: 0.5 moles / 0.08 kg = 6.25 m.

Example Problem 12: Calculating Molality with Density Given [43:28]

The video explains how to calculate the molality of a solution containing 0.85 g of ammonia in 100 ml of solution, given that the density of the solution is 1.2 g/ml. First, the number of moles of ammonia is calculated: 0.85 g / 17 g/mol (molecular mass of ammonia) = 0.05 moles. Next, the mass of the solution is calculated using the density: Mass = Density * Volume = 1.2 g/ml * 100 ml = 120 g. Then, the mass of the solvent is calculated: 120 g (solution) - 0.85 g (ammonia) = 119.15 g. This is converted to kg: 119.15 g = 0.11915 kg. Finally, the molality is calculated: 0.05 moles / 0.11915 kg = 0.42 m.