Ncert Chemistry Class 12 intext questions solutions chapter 1 Solutions

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 (C₆H₆) and carbon tetrachloride (CCl₄) 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(NO₃)₂ . 6H₂O in 4.3 L of solution (b) 30 mL of 0.5 M H₂SO₄ diluted to 500 mL.

Solution 1.3: (a) Molarity of Co(NO₃)₂ . 6H₂O in solution

Step 1: Calculate the molar mass of Co(NO₃)₂ . 6H₂O

The molar mass of Co(NO₃)₂ . 6H₂O 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(NO₃)₂ . 6H₂O

Moles of Co(NO₃)₂ . 6H₂O = \(\frac{\text{Mass}}{\text{Molar mass}}\)

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

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 (NH₂CONH₂) 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 (NH₂CONH₂) = 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: H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.

Solution 1.6: Given:

• Solubility of H₂S 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 H₂S.

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 H₂S in water

We are given the solubility of H₂S as 0.195 mol/kg, which is the molality (m). This means that there are 0.195 moles of H₂S dissolved in 1 kg of water.

To find the mole fraction ( x ), we need to compute the moles of water and the moles of H₂S.

Moles of water

1 kg of water = 1000 g
Molecular weight of water H₂O = 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 H₂O + Moles of water

Total moles = 0.195 + 55.56 = 55.755 mol

Mole fraction of H₂S

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 H₂S is 285.71 atm.

Question 1.7: Henry’s law constant for CO₂ in water is 1.67 × 10⁸ Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ 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

nH₂0= 27.78 mol of water

Now, 

Hence, quantity of CO₂ 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 (NH₂CONH₂) 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:

• P⁰ = 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, C₆H₈O₆) 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)