6 Most SAQ’s of Solutions Chapter in Inter 2nd Year Chemistry (TS/AP)

4 Marks

SAQ-1 : Define Molarity. Calculate the molarity of a solution containing 5g of NaOH in 450ml of solution.

Introduction

Molarity is a fundamental concept in chemistry, representing the concentration of a solution. It measures the number of moles of a solute within one liter of the solution.

Definition of Molarity (M)

Molarity (M) is defined as the number of moles of a solute dissolved in one liter (or 1000 milliliters) of the solution.

Formula to Calculate Molarity

$$\text{Molarity} (M) = \frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}$$

Given Data

1. Weight of NaOH (solute) = 5g
2. Molar mass (GMW) of NaOH = 40g/mol (Sodium + Oxygen + Hydrogen = 23 + 16 + 1)
3. Volume of the solution = 450 mL (or 0.450 liters)

Calculation

1. Determine the Number of Moles of NaOH:
   $$\text{Moles of NaOH} = \frac{\text{Weight of NaOH}}{\text{GMW of NaOH}} = \frac{5g}{40g/mol} = 0.125 \text{ moles}$$
2. Calculate Molarity:
   $$\text{Molarity (M)} = \frac{0.125 \text{ moles}}{0.450 \text{ liters}} = 0.2778 \text{ M}$$

Summary

The molarity of the solution containing 5g of NaOH in 450 mL of solution is approximately 0.278 M (rounded to three decimal places). This calculation indicates that the solution has 0.278 moles of NaOH per liter, illustrating the practical application of molarity in determining solution concentration.

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SAQ-2 : Define Molarity. Calculate Molality (m) of 10gm of Glucose (C₆ H₁₂ O₆) in 90gm of water.

Introduction

Molality is a crucial concept in chemistry, used to measure the concentration of a solution. It details the amount of solute present per kilogram of solvent, providing insights into the solution’s composition.

Definition of Molality (m)

Molality (m) is defined as the number of moles of the solute present in one kilogram (kg) of the solvent.

Example Calculation: Molality of a Glucose Solution

Given Values

1. Weight of Glucose (solute) = 10g
2. Molar mass of Glucose (C_6H_12O_6) = 180g/mol
3. Weight of water (solvent) = 90g

Molality Formula

$$m = \frac{\text{Number of moles of solute}}{\text{Weight of solvent in kg}}$$

$$\text{Number of moles of solute} = \frac{\text{Weight of solute}}{\text{Molar mass of solute}}$$

Calculation Steps

1. Calculate the Number of Moles of Glucose:
   $$\text{Moles of Glucose} = \frac{10g}{180g/mol} = 0.0556 \text{ moles}$$
2. Convert the Weight of Water to Kilograms:
   $$90g = 0.090 \text{ kg}$$
3. Calculate Molality:
   $$m = \frac{0.0556 \text{ moles}}{0.090 \text{ kg}} = 0.618 \text{ mol/kg}$$

Summary

The molality of the solution, containing 10g of Glucose in 90g of water, is approximately 0.618 mol/kg (rounded to three decimal places). This measure, molality, is vital for understanding solution concentration and plays a significant role in various chemical calculations and experiments.

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SAQ-3 : Define mole fraction. Calculate the mole fraction of H₂SO₄ in a solution containing 98% H₂SO₄ by mass.

Introduction

In the realm of solutions and mixtures, the mole fraction is a crucial concept that elucidates the concentration of components at a molar level, offering insights into the composition of a solution.

Definition of Mole Fraction (X)

Mole fraction (X) is defined as the ratio of the number of moles of one component to the total number of moles of all components in a solution.

Mole Fraction Formula

$$X = \frac{\text{No. of moles of one component}}{\text{Total no. of moles of all components}}$$

Example Calculation: Mole Fraction of H₂​SO₄​

Given Information

1. Mass percent of H₂​SO₄​ = 98%, implying 98g of H₂​SO₄​ in 100g of solution.
2. Mass of H₂​SO₄​ (solute) = 98g.
3. Molar mass of H_2​SO₄​ = 98g/mol.

Calculation Steps

1. Calculate the Number of Moles of H₂​SO₄​ (n_A​):
   $$n_A = \frac{98g}{98g/mol} = 1 \text{ mole}$$
2. Calculate the Mass and Moles of Water (solvent):
   • Mass of water = 2g.
   • Number of Moles of Water (n_B​): $$n_B = \frac{2g}{18g/mol} = \frac{1}{9} \text{ mole}$$
3. Calculate Mole Fraction of H₂​SO₄​ (X_A​): $$X_A = \frac{n_A}{n_A + n_B} = \frac{1}{1 + \frac{1}{9}} = \frac{9}{10} = 0.9$$

Summary

The mole fraction of H₂​SO₄​ in the solution is 0.9. This calculation demonstrates the mole fraction’s utility in quantifying the relative amounts of components in a solution, a fundamental parameter for various chemical analyses and calculations.

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SAQ-4 : What is the relative lowering of vapour pressure? How is it useful to determine the molar mass of a solute?

Introduction

Relative lowering of vapour pressure is a colligative property that describes the decrease in vapour pressure of a solvent upon the addition of a non-volatile solute. This phenomenon is directly proportional to the mole fraction of the solute in the solution.

Understanding Relative Lowering of Vapour Pressure

1. Definition: It is the ratio of the decrease in vapour pressure of the solvent (ΔP) to the vapour pressure of the pure solvent (P_0​).
2. Mathematical Expression: $$\text{Relative Lowering of Vapour Pressure} = \frac{\Delta P}{P_0} = \frac{P_0 – P}{P_0}​$$
   where P₀​ is the vapour pressure of the pure solvent, and P is the vapour pressure of the solution.

Application in Determining Molar Mass of a Solute

The relative lowering of vapour pressure can be utilized to calculate the molar mass of a solute by applying Raoult’s law. The law states that the vapour pressure of a solvent in a solution is directly proportional to the mole fraction of the solvent.

1. Raoult’s Law and Mole Fraction: $$\frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$$
   Assuming the amount of solute is much less than the amount of solvent.
2. Calculating Molar Mass:By rearranging the equation and knowing the mass of the solute (m_solute​) and solvent (m_solvent​), along with the solvent’s molar mass (M_solvent​), the molar mass of the solute (M_solute​) can be determined: $$M_{\text{solute}} = \left( \frac{\Delta P}{P_0} \right) \times \frac{M_{\text{solvent}} \times m_{\text{solute}}}{m_{\text{solvent}}}$$

Summary

Relative lowering of vapour pressure is a crucial colligative property that not only illustrates the effect of a solute on the solvent’s vapour pressure but also provides a method to calculate the molar mass of the solute. This principle is particularly useful in characterizing unknown substances and understanding solution dynamics.

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SAQ-5 : Calculate the mass of a non volatile solute (molar mass 40g mol^(-1) which should be dissolved in 114g Octane to reduce its vapour pressure to 80%.

Introduction

To determine the mass of a non-volatile solute required to reduce the vapour pressure of Octane to 80% of its original value, we can use the concept of relative lowering of vapour pressure.

Given Data

1. Molar Mass of Solute: 40 g mol⁻¹
2. Mass of Octane (Solvent): 114g
3. Reduction in Vapour Pressure: To 80% of original

Steps to Calculate the Mass of the Solute

1. Relative Lowering of Vapour Pressure: $$\frac{\Delta P}{P_0} = 1 – \frac{P}{P_0} = 1 – 0.80 = 0.20$$
2. Use of Raoult’s Law: According to Raoult’s law, the relative lowering of vapour pressure is equal to the mole fraction of the solute (X_solute​).
   $$\frac{\Delta P}{P_0} = X_{\text{solute}}$$
3. Calculating Mole Fraction of Solute: Since the relative lowering of vapour pressure is 0.20, X_solute​=0.20.
4. Moles of Octane: Molar mass of Octane (C₈​H₁₈​) = 114g mol⁻¹
   $$\text{Moles of Octane} = \frac{\text{Mass of Octane}}{\text{Molar Mass of Octane}} = \frac{114 \, \text{g}}{114 \, \text{g mol}^{-1}} = 1 \, \text{mol}$$
5. Calculating Moles of Solute: Using the mole fraction and the moles of octane, we can calculate the moles of solute:
   $$X_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{octane}}}$$
   Solving for n_solute​, we find:
   $$n_{\text{solute}} = X_{\text{solute}} \times n_{\text{octane}} = 0.20 \times 1 \, \text{mol} = 0.20 \, \text{mol}$$
6. Calculating Mass of Solute:
   $$\text{Mass of Solute} = n_{\text{solute}} \times \text{Molar Mass of Solute} = 0.20 \, \text{mol} \times 40 \, \text{g mol}^{-1} = 8 \, \text{g}$$

Summary

To reduce the vapour pressure of 114 g of Octane to 80% of its original value, 8 g of a non-volatile solute with a molar mass of 40 g/mol must be dissolved in it. This calculation demonstrates the application of the relative lowering of vapour pressure in determining the amount of solute required for a desired vapour pressure reduction.

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SAQ-6 : The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5g when added to 39.0 g of benzene (molar mass 78 g mol^(-1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance?

Introduction

This calculation involves determining the molar mass of a non-volatile solute by observing the change in vapour pressure of a solution compared to pure benzene.

Given Data

1. Vapour Pressure of Pure Benzene: 0.850 bar
2. Vapour Pressure of Solution: 0.845 bar
3. Mass of Solute: 0.5g
4. Mass of Benzene (Solvent): 39.0g
5. Molar Mass of Benzene: 78g mol⁻¹

Calculation Steps

1. Calculate the Relative Lowering of Vapour Pressure:
   $$\frac{\Delta P}{P_0} = \frac{P_0 – P}{P_0} = \frac{0.850 – 0.845}{0.850} = 0.0059$$
2. Calculate Moles of Benzene:
   $$\text{Moles of Benzene} = \frac{\text{Mass of Benzene}}{\text{Molar Mass of Benzene}} = \frac{39.0}{78} = 0.5 \, \text{mol}$$
3. Apply Raoult’s Law to Determine Moles of Solute: Raoult’s law states that the relative lowering of vapour pressure is equal to the mole fraction of the solute. Therefore, we can express the number of moles of solute (n_solute​) in terms of the relative lowering of vapour pressure and moles of solvent (n_solvent​):
   $$\frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}}$$
   Given that the solute concentration is low, n_solute​ is much smaller than n_solvent​, simplifying to:
   $$\frac{\Delta P}{P_0} \approx \frac{n_{\text{solute}}}{n_{\text{solvent}}}$$
   $$n_{\text{solute}} = \frac{\Delta P}{P_0} \times n_{\text{solvent}} = 0.0059 \times 0.5 = 0.00295 \, \text{mol}$$
4. Calculate Molar Mass of Solute:
   $$\text{Molar Mass of Solute} = \frac{\text{Mass of Solute}}{\text{Moles of Solute}} = \frac{0.5}{0.00295} = 169.49 \, \text{g mol}^{-1}$$

Summary

The molar mass of the solid substance added to benzene is approximately 169.49 g/mol. This calculation demonstrates the utility of vapour pressure measurements and the relative lowering of vapour pressure concept in determining the molar mass of a non-volatile solute in a solution.