## Solutions intext questions solutions: Class 12 Chemistry Chapter 1 intext questions solutions

Textbook | Ncert |

Class | Class 12 |

Subject | Chemistry |

Chapter | Chapter 1 |

Chapter Name | Solutions Class 12 intext questions solutions |

Category | Intext questions |

Medium | English |

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### Question 1.1:** **Calculate the mass percentage of benzene (C_{6}H_{6}) and carbon tetrachloride (CCl_{4}) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.

**Solution 1.1:** To calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) in the solution, we use the formula for mass percentage:

Mass percentage of a component = \(\frac{\text{Mass of the component}}{\text{Total mass of the solution}} \times 100\)

**Calculate the total mass of the solution**

Total mass of the solution = Mass of benzene + Mass of carbon tetrachloride

Total mass of the solution = 22 g + 122 g = 144 g

**Calculate the mass percentage of benzene**

Mass percentage of benzene = \(\frac{22 \, \text{g}}{144 \, \text{g}} \times 100 = 15.28\%\)

**Calculate the mass percentage of carbon tetrachloride**

Mass percentage of carbon tetrachloride = \(\frac{122 \, \text{g}}{144 \, \text{g}} \times 100 = 84.72\%\)

- Mass percentage of carbon tetrachloride (CCl₄) =
**84.72%** - Mass percentage of benzene (C₆H₆) =
**15.28%**

### Question 1.2:** **Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.

**Solution 1.2:** To calculate the mole fraction of benzene in a solution where the mass percentage of benzene is 30% and the solvent is carbon tetrachloride (CCl₄), we follow these steps:

**Define the masses of benzene and carbon tetrachloride**

Since the solution contains 30% by mass of benzene, we can assume we have 100 g of the solution for simplicity. Thus:

- Mass of benzene (C₆H₆) = 30 g
- Mass of carbon tetrachloride (CCl₄) = 70 g

**Calculate the molar masses**

The molar masses of benzene and carbon tetrachloride are calculated as follows:

- Benzene (C₆H₆): 12.01 × 6 + 1.008 × 6 = 78.11 g/mol
- Carbon tetrachloride (CCl₄): 12.01 × 1 + 35.45 × 4 = 153.81 g/mol

**Calculate the number of moles of each component**

Moles of benzene = \(\frac{\text{Mass of benzene}}{\text{Molar mass of benzene}} \)

\(= \frac{30 \, \text{g}}{78.11 \, \text{g/mol}} = 0.384 \, \text{mol}\)

Moles of carbon tetrachloride = \(\frac{\text{Mass of carbon tetrachloride}}{\text{Molar mass of carbon tetrachloride}}\)

\( = \frac{70 \, \text{g}}{153.81 \, \text{g/mol}} = 0.455 \, \text{mol}\)

**Calculate the total moles in the solution**

Total moles = 0.384 mol + 0.455 mol = 0.839 mol

**Calculate the mole fraction of benzene**

The mole fraction of benzene \(( X_{\text{benzene}} )\) is given by:

X_{benzene} = \(\frac{\text{Moles of benzene}}{\text{Total moles}}\)

\( = \frac{0.384}{0.839} = 0.458\)

The mole fraction of benzene in the solution is **0.458**.

### Question 1.3:** **Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO_{3})_{2 }. 6H_{2}O in 4.3 L of solution (b) 30 mL of 0.5 M H_{2}SO_{4 }diluted to 500 mL.

**Solution 1.3:** (a) Molarity of Co(NO_{3})_{2 }. 6H_{2}O in solution

**Step 1: Calculate the molar mass of Co(NO _{3})_{2} . 6H_{2}O**

The molar mass of Co(NO_{3})_{2 }. 6H_{2}O is calculated as follows:

- Cobalt (Co): \( 1 \times 58.93 \, \text{g/mol} \)
- Nitrogen (N): \( 2 \times 14.01 \, \text{g/mol} \)
- Oxygen (O): \( 6 \times 16.00 \, \text{g/mol} \) (from 2 nitrate groups)
- Water (H₂O): \( 6 \times (2 \times 1.008 + 16.00) \, \text{g/mol} \)

So:

Molar mass = 58.93 + (2 × 14.01) + (6 × 16.00) + 6 × (2 × 1.008 + 16.00) = 291.03 g/mol

**Step 2: Calculate the number of moles of Co(NO _{3})_{2} . 6H_{2}O**

Moles of Co(NO_{3})_{2 }. 6H_{2}O = \(\frac{\text{Mass}}{\text{Molar mass}}\)

\( = \frac{30 \, \text{g}}{291.03 \, \text{g/mol}} = 0.103 \, \text{mol}\)

**Step 3: Calculate the molarity**

Molarity (M) is defined as:

Molarity = \(\frac{\text{Moles of solute}}{\text{Volume of solution in liters}}\)

Given that the volume of the solution is 4.3 L:

Molarity = \(\frac{0.103 \, \text{mol}}{4.3 \, \text{L}} = 0.024 \, \text{M}\)

So, the molarity of \( \text{Co(NO}_3\text{)}_2 \cdot 6\text{H}_2\text{O} \) is **0.024 M**

(b) Molarity of H₂SO₄ after dilution

**Step 1: Use the dilution formula**

The dilution formula is:

\(M_1 V_1 = M_2 V_2\)

Where:

- \( M_1 \) = initial molarity of the solution (0.5 M)
- \( V_1 \) = initial volume of the solution (30 mL = 0.030 L)
- \( M_2 \) = final molarity after dilution (to be calculated)
- \( V_2 \) = final volume of the solution (500 mL = 0.500 L)

**Step 2: Solve for \( M_2 \)**

\(M_2 = \frac{M_1 V_1}{V_2} \)

\(= \frac{(0.5 \, \text{M}) \times (0.030 \, \text{L})}{0.500 \, \text{L}} \)

= 0.03 M

So, the molarity of the diluted H₂SO₄ solution is **0.03 M**.

### Question 1.4:** **Calculate the mass of urea (NH_{2}CONH_{2}) required in making 2.5 kg of 0.25 molal aqueous solution.

**Solution 1.4:** Moles of urea = 0.25 mol

Mass of solvent (water) = 1 kg = 1000 g

Molar mass of urea (NH_{2}CONH_{2}) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16

= 60 g mol^{−1}

∴ Mass of urea in 1000 g of water = 0.25 mol × 60 g mol^{−1}

= 15 g

Total mass of solution = 1000 + 15 g

= 1015 g

= 1.015 kg

Thus, 1.015 kg of solution contain urea = 15 g

∴ 2.5 kg of solution will require urea = \(\frac{15 \, \text{g}}{1.015 \, \text{Kg}} \times 2.5kg\)

= 36.95 g

= 37 g (approximately)

Hence, the mass of urea required is 37 g.

### Question 1.5:** **Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL^{−1}.

**Solution 1.5:** Given:

- 20% (mass/mass) aqueous KI solution
- Density of solution = 1.202 g/mL

Assumptions:

- 20% (mass/mass) means 20 g of KI is present in 100 g of solution.

**Part (a) Molality (m):**

**Molality** is defined as the number of moles of solute per kilogram of solvent.

Step 1: Calculate moles of KI

Molecular weight of KI = 39 (K) + 127 (I) = 166 g/mol

Given mass of KI = 20 g

Moles of KI = \(\frac{20 \, \text{g}}{166 \, \text{g/mol}} = 0.1205 \, \text{mol}\)

Step 2: Calculate the mass of water (solvent)

Mass of the solution = 100 g

Mass of KI = 20 g

Mass of water = 100 g – 20 g = 80 g = 0.080 kg

Step 3: Calculate molality

Molality = \(\frac{\text{Moles of KI}}{\text{Mass of solvent (kg)}}\)

\( = \frac{0.1205 \, \text{mol}}{0.080 \, \text{kg}} = 1.506 \, \text{mol/kg}\)

**Part (b) Molarity (M):**

**Molarity** is defined as the number of moles of solute per liter of solution.

Step 1: Calculate the volume of the solution

We know the mass of the solution = 100 g, and the density of the solution = 1.202 g/mL.

Volume of solution = \(\frac{\text{Mass of solution}}{\text{Density}} \)

\(= \frac{100 \, \text{g}}{1.202 \, \text{g/mL}} = 83.195 \, \text{mL} = 0.0832 \, \text{L}\)

Step 2: Calculate molarity

Molarity = \(\frac{\text{Moles of KI}}{\text{Volume of solution (L)}}\)

\( = \frac{0.1205 \, \text{mol}}{0.0832 \, \text{L}} = 1.45 \, \text{mol/L}\)

Part (c) Mole fraction of KI:

**Mole fraction** is defined as the ratio of the number of moles of the solute to the total number of moles of all components.

Step 1: Calculate moles of water

Molecular weight of water = 18 g/mol

Moles of water = \(\frac{80 \, \text{g}}{18 \, \text{g/mol}} = 4.444 \, \text{mol}\)

Step 2: Calculate the mole fraction of KI

Mole fraction of KI = \(\frac{\text{Moles of KI}}{\text{Moles of KI + Moles of water}}\)

\( = \frac{0.1205}{0.1205 + 4.444} \)

\(= \frac{0.1205}{4.5645} = 0.0263\)

- (a)
**Molality**= 1.506 mol/kg - (b)
**Molarity**= 1.45 mol/L - (c)
**Mole fraction of KI**= 0.0263

### Question 1.6:** **H_{2}S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H_{2}S in water at STP is 0.195 m, calculate Henry’s law constant.

**Solution 1.6:** Given:

- Solubility of H
_{2}S in water at STP = 0.195 molality (mol/kg) - STP (Standard Temperature and Pressure):
- Temperature ( T ) = 273.15 K
- Pressure ( P ) = 1 atm

We need to calculate Henry’s law constant k_{H} for H_{2}S.

**Understand Henry’s Law**

Henry’s law states:

\(P = k_H \cdot x\)

Where:

- P = partial pressure of the gas (1 atm at STP)
- k
_{H}= Henry’s law constant (to be calculated) - x = mole fraction of the gas in solution

**Calculate mole fraction of H _{2}S in water**

We are given the solubility of H_{2}S as 0.195 mol/kg, which is the molality (m). This means that there are 0.195 moles of H_{2}S dissolved in 1 kg of water.

To find the mole fraction ( x ), we need to compute the moles of water and the moles of H_{2}S.

**Moles of water**

1 kg of water = 1000 g

Molecular weight of water H_{2}O = 18 g/mol

Moles of water = \(\frac{1000 \, \text{g}}{18 \, \text{g/mol}} \)\(= 55.56 \, \text{mol}\)

Total moles of solution

Total moles = Moles of H_{2}O + Moles of water

Total moles = 0.195 + 55.56 = 55.755 mol

Mole fraction of H_{2}S

x = \(\frac{\text{Moles of } \text{H}_2\text{S}}{\text{Total moles}}\)

\( = \frac{0.195}{55.755} = 0.0035\)

Calculate Henry’s law constant k_{H}

Using Henry’s law equation \( P = k_H \cdot x \), we can solve for k_{H}:

k_{H} = \(\frac{P}{x} \)

\(= \frac{1 \, \text{atm}}{0.0035} \)

\(= 285.71 \, \text{atm}\)

The Henry’s law constant for H_{2}S is **285.71 atm**.

### Question 1.7:** **Henry’s law constant for CO_{2} in water is 1.67 × 10^{8} Pa at 298 K. Calculate the quantity of CO_{2} in 500 mL of soda water when packed under 2.5 atm CO_{2 }pressure at 298 K.

**Solution 1.7:** It is given that:

K_{H} = 1.67 × 10^{8 Pa}

= 2.5 atm = 2.5 × 1.01325 × 10^{5 Pa}

= 2.533125 × 10^{5 Pa}

According to Henry’s law:

= 0.00152

We can write,

[Since, is negligible as compared to]

In 500 mL of soda water, the volume of water = 500 mL

[Neglecting the amount of soda present]

We can write:

500 mL of water = 500 g of water

nH_{2}0= 27.78 mol of water

Now,

Hence, quantity of CO_{2} in 500 mL of soda water = (0.042 × 44)g

= 1.848 g

### Question 1.8:** **The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K . Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

**Solution 1.8:** Given data:

- Vapor pressure of pure liquid A \(( P_A^0 )\) = 450 mm Hg
- Vapor pressure of pure liquid B \(( P_B^0 )\) = 700 mm Hg
- Total vapor pressure of the mixture \(( P_{\text{total}} )\) = 600 mm Hg

Let \( x_A \) and \( x_B \) be the mole fractions of liquids A and B , respectively, in the liquid phase.

According to Raoult’s Law:

\(P_{\text{total}} = P_A + P_B \)\(= x_A P_A^0 + x_B P_B^0\)

Where P_{A} and P_{B} are the partial pressures of liquids A and B in the mixture.

Since \( x_A + x_B = 1 \), we can express \( x_B \) as \( x_B = 1 – x_A \).

Now, substitute this into the total pressure equation:

600 = \(x_A \times 450 + (1 – x_A) \times 700\)

Solve for ( x_{A} ).

Expand and solve:

\(600 = 450 x_A + 700 – 700 x_A\)

\(600 = 700 – 250 x_A\)

\(250 x_A = 700 – 600\)

\(250 x_A = 100\)

\(x_A = \frac{100}{250} = 0.4\)

Thus, the mole fraction of ( A ) in the liquid phase is ( x_{A} = 0.4 ).

Since ( x_{A} + x_{B }= 1 ), the mole fraction of ( B ) is: x_{B} = 1 – 0.4 = 0.6

**Find the composition of the vapor phase.**

For the vapor phase, the mole fractions of ( A ) and ( B ) in the vapor phase (denoted as y_{A} and y_{B} ) are determined using Dalton’s law:

\(y_A = \frac{P_A}{P_{\text{total}}} =\) \( \frac{x_A \times P_A^0}{P_{\text{total}}}\)

\(y_B = \frac{P_B}{P_{\text{total}}} \) \(= \frac{x_B \times P_B^0}{P_{\text{total}}}\)

Now calculate each:

- Partial pressure of ( A ):

\(P_A = x_A \times P_A^0 = 0.4 \times 450\) = 180 mm Hg - Partial pressure of ( B ):

\(P_B = x_B \times P_B^0 = 0.6 \times 700 \) = 420 mm Hg

Now, calculate ( y_{A} ) and ( y_{B} ):

\(y_A = \frac{P_A}{P_{\text{total}}} = \frac{180}{600} = 0.3\)

\(y_B = \frac{P_B}{P_{\text{total}}} = \frac{420}{600} = 0.7\)

Final Answer:

- The composition of the liquid phase is:
- Mole fraction of A \(( x_A )\) = 0.4
- Mole fraction of B \(( x_B )\) = 0.6

- The composition of the vapor phase is:
- Mole fraction of A \(( y_A )\) = 0.3
- Mole fraction of B \(( y_B )\) = 0.7

### Question 1.9:** **Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH_{2}CONH_{2}) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

**Solution 1.9:** To solve this problem, we will use the concept of **Raoult’s Law** for non-volatile solutes. The lowering of vapor pressure when a non-volatile solute (urea) is added to a solvent (water) can be calculated using the formula for **relative lowering of vapor pressure**:

\(\frac{P^0 – P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}\)

Where:

- P
^{0}= Vapor pressure of pure water - P = Vapor pressure of the solution
- n
_{solute}= Moles of the solute (urea) - n
_{solvent}= Moles of the solvent (water)

Given data:

- Vapor pressure of pure water \(( P^0 )\) = 23.8 mm Hg
- Mass of urea = 50 g
- Mass of water = 850 g
- Molar mass of urea (NH₂CONH₂) = 14 + 2 × 1 + 12 + 16 + 14 + 2 × 1 = 60 g/mol
- Molar mass of water (H₂O) = 18 g/mol

**Calculate the moles of urea (n _{solute} ).**

n_{solute} = \(\frac{\text{mass of urea}}{\text{molar mass of urea}}\)

\(= \frac{50}{60} = 0.833 \, \text{mol}\)

**Calculate the moles of water (n _{solute} ).**

n_{solute} = \(\frac{\text{mass of water}}{\text{molar mass of water}}\)

\( = \frac{850}{18} = 47.22 \, \text{mol}\)

**Calculate the relative lowering of vapor pressure.**

Using the formula for relative lowering of vapor pressure:

\(\frac{P^0 – P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}\)

\(\frac{23.8 – P}{23.8} = \frac{0.833}{47.22}\)

\(\frac{23.8 – P}{23.8} = 0.01764\)

Solve for ( P ) (vapor pressure of the solution).

Now, solve for ( P ):

\(23.8 – P = 23.8 \times 0.01764 = 0.4196\)

\(P = 23.8 – 0.4196 = 23.4 \, \text{mm Hg}\)

Calculate the relative lowering of vapor pressure.

The relative lowering of vapor pressure is:

Relative lowering

= \(\frac{P^0 – P}{P^0} \)

\(= \frac{23.8 – 23.38}{23.8} \)

\(= \frac{0.4196}{23.8} = 0.01764\)

The vapor pressure of the water in the solution is **23.4 mm Hg**.

The relative lowering of vapor pressure is **0.01764** or **1.764%**.

### Question 1.10:** **Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.

**Solution 1.10:** This problem involves the concept of **boiling point elevation**, which occurs when a non-volatile solute (sucrose) is added to a solvent (water). The formula for boiling point elevation is given by:

\(\Delta T_b = K_b \cdot m\)

Where:

- \( \Delta T_b \) is the boiling point elevation.
- \( K_b \) is the ebullioscopic constant (boiling point elevation constant) of the solvent.
- m is the molality of the solution.

**Given data**

- Boiling point of pure water at 750 mm Hg = 99.63°C.
- Boiling point of the solution = 100°C.
- Mass of water = 500 g = 0.5 kg.
- Molar mass of sucrose \(( C_{12}H_{22}O_{11} )\) = 12 × 12 + 22 × 1 + 11 × 16 = 342 g/mol.
- Boiling point elevation constant for water \(( K_b )\) = 0.52°C·kg/mol (for water).

**Calculate the boiling point elevation \(( \Delta T_b )\).**

The boiling point elevation is the difference between the boiling point of the solution and the boiling point of pure water at the given pressure:

\(\Delta T_b = 100 – 99.63 = 0.37°C\)

**Calculate the molality of the solution.**

Using the boiling point elevation formula:

\(\Delta T_b = K_b \cdot m\)

\(0.37 = 0.52 \cdot m\)

\(m = \frac{0.37}{0.52} = 0.7115 \, \text{mol/kg}\)

**Calculate the moles of sucrose required.**

Molality is defined as moles of solute per kilogram of solvent, so:

m = \(\frac{\text{moles of sucrose}}{\text{mass of solvent (in kg)}}\)

\(0.7115 = \frac{\text{moles of sucrose}}{0.5}\)

moles of sucrose = \(0.7115 \times 0.5 = 0.35575 \, \text{mol}\)

**Calculate the mass of sucrose required.**

Now, using the molar mass of sucrose (342 g/mol):

mass of sucrose = moles of sucrose × molar mass of sucrose

mass of sucrose = \(0.35575 \times 342 = 121.67 \, \text{g}\)

To raise the boiling point of 500 g of water to 100°C, approximately **121.67 g of sucrose** needs to be added.

### Question 1.11: Calculate the mass of ascorbic acid (Vitamin C, C_{6}H_{8}O_{6}) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. K_{f} = 3.9 K kg mol^{−1}.

**Solution 1.11:** To calculate the mass of ascorbic acid (Vitamin C, C₆H₈O₆) required to lower the melting point of 75 g of acetic acid by 1.5°C, we use the formula for freezing point depression:

\(\Delta T_f = K_f \times m\)

Where:

- \(\Delta T_f = 1.5^\circ C\) (freezing point depression),
- K
_{f }= 3.9 K kg/mol (freezing point depression constant of acetic acid), - m is the molality of the solution (moles of solute per kg of solvent).

**Rearrange to solve for molality (m):**

\(m = \frac{\Delta T_f}{K_f} \)

\(= \frac{1.5}{3.9} \approx 0.3846 \text{ mol/kg}\)

**Calculate the moles of ascorbic acid needed:**

moles = m × kg of solvent

\(= 0.3846 \times \frac{75}{1000} = 0.0288 \text{ mol}\)

Calculate the mass of ascorbic acid using its molar mass ( C₆H₈O₆, molar mass = 176 g/mol):

mass = moles × molar mass

\(= 0.0288 \times 176 = 5.07 \text{ g}\)

So, 5.07 g of ascorbic acid is needed.

### Question 1.12: Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

**Solution 1.12:** It is given that:

Volume of water, V = 450 mL = 0.45 L

Temperature, T = (37 + 273)K = 310 K

Number of moles of the polymer,

We know that:

Osmotic pressure,

= 30.98 Pa

= 31 Pa (approximately)