NCERT Solutions for Class 12 Chemistry Chapter 1 – Solutions

Intext Questions with Solutions of Class 12 Chemistry Chapter 1 – Solutions

1.1

Ans – The mass percentage is the ratio between the solute’s mass and the total mass of the solution, multiplied by 100.

The mass proportion of each component is provided as follows;

The solution consists just in two elements: C₆H₆ and CCl₄.
Consequently, the mass percentage of CCl₄ is computed as 100 – 15.28, thereby obtaining 84.72%.
C₆H₆ has mass percentages of 15.28% and CCl₄ of 84.72%. respectively.

1.2

Ans – The mole fraction is the ratio of the moles of solute to the total moles of the solution.

This solution has 30% by mass of benzene in carbon tetrachloride, indicating that 30 g of benzene is contained in 100 g of the solution. Therefore, carbon tetrachloride constitutes 70 g inside 100 g of the solution.
It is known that,

⇒ No. of moles of benzene = 30/78 = 0.385mol
⇒ No. of moles of carbon tetrachloride = 70/154 = 0.455mol

∴ The mole fraction of benzene (C₆H₆) in the solution = 0.458.

1.3

Ans – a) Molarity is defined as the quantity of moles of solute per litre of solution.

Here V_solution is given as 4.3L
The molar mass of Co(NO₃)₂.6H₂O is 310.7 g/mol.
Number of moles = 30/310.7 = 0.0966
Therefore, molarity = 0.0966/4.3 = 0.022 M

b) The molarity of 30 mL of 0.5 M H2SO4 diluted to 500 mL.
There are 0.5 moles of H2SO4 in 1000 ml of 0.5 M H2SO4.
Consequently, 30 ml of 0.5 M H₂SO₄ has 0.5/1000 × 30 = 0.015 moles of H₂SO₄.
The volume of the solution is 500 millilitres, equivalent to 0.5 litres.
This means that the molarity is 0.015/0.5, which is 0.03 M.

1.4

Ans – Molality is defined as the ratio of moles of solute to grammes of solvent mass.

A 0.25 molal solution of urea signifies that 0.25 moles of urea are dissolved in 1000 grammes of solvent.
The molar mass of the solute, urea, is 60 g/mol.
Hence, the mass of urea in the solution is 15 grammes, calculated as 0.25 multiplied by 60.
Total mass of solution = 1000 g solvent + 15 g solute = 1015 g equals 1.015 kg.
This indicates that, a 1.015 kg solution contains 15 g of urea.
As a result, 2.5 kg solution will contain 37 g of urea, calculated as follows: 
15/1.015 x 2.5 = 37.
∴ the requisite mass of urea will be 37 g.

1.5

Ans – a) In a 20% aqueous KI solution, there are 20 grammes of KI in 100 grammes of solution.
Mass of water (solvent) = 100 – 20 = 80 g.
The solution’s volume is given as follows:

Potassium iodide’s (KI) molar mass is 166.0028 g/mol.

Water has a molar mass of 18 g/mol.

b)

c)

1.6

Ans – The molality of H₂S in water is 0.195 m, indicating that there are 0.195 moles of H₂S in 1 kg of water (1000 g).
Molar mass of water = 18 g/mol.
The quantity of moles of water is calculated as 1000 divided by 18, resulting in 55.55 moles.
The mole fraction of H₂S is:

Pressure at standard temperature and pressure (STP) = 0.987 bar
∴ Using Henry’s law, we obtain:

Thus, Henry’s constant will equal 282.80 bar.

1.7

Ans –

500 ml of soda water is equivalent to 500 ml of water, which indirectly indicates 500 grammes of the same.
Molar mass of water = 18 g/mol.

No. of moles of CO₂ will be,

Molar mass of CO₂ = 44 g/mol.

1.8

Ans – The vapour pressures of the specified pure liquids are as follows:
P_A⁰ = 450mm Hg
P_B⁰ = 700mm Hg
P_Total = 600mm Hg

According to Raoult’s law we know,
P_Total = P_A + P_B
= X_AP_A⁰ + X_BP_B⁰
= X_AP_A⁰ + (1 – X_A)P_B⁰
= P_B⁰ + (P_A⁰ – P_B⁰)X_A
∴ 600 = 700 + (450 − 700)X_A
⇒ X_A = 0.4
⇒ X_B = 1 – 0.4 = 0.6

P_A = X_AP_A⁰
= 0.4 × 450 =180mm Hg
P_B = X_BP_B⁰
= 0.6 × 700 = 420mm Hg
In the vapor phase,

1.9

Ans – The vapour pressure of pure water, P⁰, is 23.8 mm Hg.
The mass of urea, W₂, is 50 g
The molar mass of urea, M₂, is 60 g/mol.
Mass of water, W₁ = 850 g
Molar mass of water = 18 g/mol.

Applying Raoult’s law,

Putting P⁰ = 23.8 mm, we will get,

1.10

Ans – Water’s boiling point = 100°C.
Water’s boiling point at 750 mm Hg = 99.63°C.
Boiling point increase, ΔT_b = 100 – 99.63 = 0.37
The ebullioscopic constant, K_b = 0.52.
Mass of water, W₁ = 500 g
Molar mass of water = 18 g/mol.
The molar mass of sucrose, M₂ = 342 g/mol.
The rise in boiling point is specified as;
ΔT_b = K_b x m

∴ 121.67 g of sucrose is to be dissolved in water to achieve a boiling point of 100°C.

1.11

Ans – Lowering in freezing point, ΔT_f = 1.5^∘ C
Cryoscopic constant, K_f = 3.9Kg/mol
Mass of acetic acid, W₁ = 75 g
Molar mass of acetic acid, M₁ = 60 g/mol
Molar mass of ascorbic acid, M₂ = 176 g/mol
Mass of ascorbic acid, W₂ = ?
The lowering in freezing point is given as;

Ascorbic acid has a mass of 5.077 g.

1.12

Ans –

Now, applying the osmotic pressure formula,

Exercise Questions with Solutions of Class 12 Chemistry Chapter 1 – Solutions

1.1

Ans – A homogeneous combination of over 2 chemically non-reacting elements is defined as a solution. 9 different variants of solutions are deemed under the 3 major states: Gaseous, Liquid, and Solid.

Here are the various types of solutions with desired examples –

• Solid in Solid: Ornaments like Cu/Ag with Au.
• Gas in Gas: Air, The combination of O₂ & N₂ elements.
• Liquid in Gas: Water vapor is a relevant example of this variant.
• Solid in Gas: Smoke, Camphor vapors in N₂ gas.
• Gas in Liquid: Aerated water, O₂ dissolved in H₂O.
• The liquid in Liquid: Vinegar solution is a prime example of this variant.
• Solid in Liquid: Glucose is dissolved in water and saline water.
• Gas in Solid: Solution of hydrogen in platinum.
• Liquid in Solid: Amalgams such as Mg and Hg.

1.2

Ans – We all know that a gaseous state is defined as a solute in a solid solution, which is also referred to as a gas-solid solution. Some of the well-known examples are solid CO­­₂ which is commonly seen in fire extinguishers, combined solution of H₂ in palladium, multiple dissolved gases in earth-based minerals, and the list goes on.

1.3

Ans – (i) Mole fraction can be defined as the overall ratio of moles present in a solute when compared to the total moles present in a solution.

(ii) Molality is defined as the ratio of solute moles per gram of solvent mass.

It is a much better and more reliable option to showcase the overall concentration of the total solute since this does not affect the modifications in temperature compared to molarity (based on the total volume of the solution).

(iii) Molarity is the overall no. of moles of solute available per litre of a solution.

(iv) Mass percentage is the total ratio of the mass of the solute that gets separated by the total mass of the solution which is then multiplied by 100.

1.4

Ans – As we all know, we already contain a 68% nitric acid regarding mass in an aqueous solution where 68 grams of nitric acid is readily available in 100 grams of solution.
Then, the overall molar mass of nitric acid remains at 63 gms/mol
Hence, the no. of moles of nitric acid can be given as, 68/63 = 1.079mol
So, the total density of the solution stands at 1.504 gms/ml.
Finally,

Similarly, the molarity can be defined as,

1.5

Ans – As we already know, a 10% w/w mixture of sugar in water, which is equivalent to 10 g of glucose in 90 grams of liquid, is what we have provided.
As per our recent understanding;
Glucose molar mass in total remains 180 gms/mol
The water’s molar mass stands at 18 g/mol.
Therefore, 10/180 = 0.055 mol is the number of moles of glucose in the water solution.
90/18 = 5mol is the number of moles of water in the solution.
Considering the values mentioned previously;
The Molality can be provided as,

Any element’s molecular percentage shall be expressed as follows:

We are already been informed that the overall density of the solution remains at 1.2 gms/ml. So,

1.6

Ans – As we already know,
Assume for the moment that we are dealing with x g of Na₂CO₃ in an amalgam of 1 g. Consequently, (1 – x) g of NaHCO₃ will be in the same.
We are aware that,
The total molar mass of Na₂CO₃ = 106 gms/mol
The total molar mass of NaHCO₃ = 84 gms/mol

Since we now know that they are equimolar,

Thus,

We must examine the effects for each to determine the amount of millilitres of 0.1 M HCl needed to fully interact with a 1 g combination of Na₂CO₃ & NaHCO₃.

From this we get that each mole of Na₂CO₃ requires 2 moles of HCl.
⇒ 0.00526 moles of Na₂CO₃ requires 0.00526 x 2 = 0.01052 moles of HCl

From this we get that each mole of NaHCO₃ requires a mole of HCl.
⇒ 0.00503 moles of NaHCO₃ requires 0.00503 moles of HCl
Thus, total moles of HCl required = 0.01052 + 0.00503 = 0.01555 moles
There are 1.1 moles of 0.1 M HCl in a 1000 ml solution.

Thus, the volume necessary for complete reaction with the mixture is 155.5 ml.

1.7

Ans – As we already know,
The total weight of the solute is divided by the mass of the solution itself, multiplied by 100. Therefore, the total percentage value of mass is defined as,

A combination of 400 g of 40% solutions by mass & 300 g of 25% solution is provided here. As an outcome, the final solution will contain the following:
Solute total mass = 75 + 160 = 235 g
The solution’s total weight is 300 + 400 = 700 g.
Consequently, the mass percentage is provided as
235/700 * 100 = 33.57% is the mass proportion of solute present in the desired solution. Likewise, 100 – 33.57 = 66.42% is the mass proportion of water in the solution.

1.8

Ans –

Thus,

1.9

Ans – (i) We are aware that the value of measurement might indicate the quantity in parts per million (ppm) or parts in 106106 of the mixture.
Meanwhile, 15 ppm in this context refers to 15 parts per 10⁶ of solution.
As a result, the mass proportion is provided as,

(ii) 15 ppm indicates that there is 15 g of chloroform in 10⁶ grams of solution, meaning that the solvent density is equal to 10⁶ grams.
Chloroform’s mole fraction is 119.5 grams/mol.

1.10

Ans – Both of these substances have a high propensity to create intermolecular hydrogen bonds. Alcohol & H₂O molecules establish H-bonds if these two liquids are mixed, resulting in the creation of a solution. Compared to pure H₂O, these connections are weaker & less widespread. As a result, they exhibit a favourable departure from optimum conduct. Consequently, compared to pure water & pure alcohol, the normal versions of the alcohol and water combination will possess a reduced boiling point & a larger vapor pressure.

1.11

Ans – Exothermic means that surplus thermal energy is released during the dissolution of gasses in water. The state of equilibrium moves backward following Le Chatelier’s principle if the operation’s temperature rises more. As a result, gasses lose their solubility in liquids.

1.12

Ans – As Henry’s law states that:
The partial pressure of a gas is entirely linked to its solubility in a liquid at a certain temperature level. This law governs how pressure affects the solubility of a gas element that is already present in a liquid state.
In mathematics, P = K_Hx
wherein P is defined as the partial pressure of the gas element.
The molecular proportion of gas present in the mixture is denoted by x.
Henry’s Law constant is K_H.
Some common applications of Henry’s law include:

1. Plunging into the sea.
2. In the manufacturing of fizzy drinks.

1.13

Ans – Using Henry’s law, we all can understand that the overall solubility of a gas in a liquid medium can be directly proportional to the overall pressure of the gas. This nominal proportionality sign can be swiftly replaced using Henry’s constant. Hence, it can be defined as;
m = K_H × P

Therefore, the mole fraction can be directly proportional to the overall ethane mass.
Hence,
In case 1 we can write,
6.56 × 10⁻³ = K_H × 1 
In case 2 we can write,
5 × 10⁻² = K_H × x, where x is the partial pressure of the gas when the solution comprises 5 × 10⁻²g of ethane.
Now, equating both the equations together we get,

Therefore, partial pressure of the gas will be, x = 7.62 bar

1.14

Ans – Going by Raoult’s law, the positive deviation can be depicted as,
Solutions that showcase positive deviation from Raoult’s law if they contain higher vapor pressure well above the prescribed limit as mentioned in the law.
Meanwhile, Δ_mixH remains positive since the energy can be gathered for dismantling the robust connection and building a feeble connection. Similarly, Δ_mixV stands in the positive zone based on the expansion of volume that occurs during this event.
Going by Raoult’s law, the negative deviation can be depicted as,
Solutions that portray negative deviation using Raoult’s law if they contain lower vapor pressure well above the prescribed limit as mentioned in the law.
Meanwhile, Δ_mixH stays at the negative zone since the energy is dispersed because of the external replacement of feeble interactions shadowed by the much more robust ones.

1.15

Ans – As we all know,
The vapor pressure of pure water during its boiling point is defined as P⁰ = 1.103 bar.
Meanwhile, the vapor pressure of the solution can be calculated as P_s = 1.004 bar.
Based on the above statement, an aqueous solution has a 2% non-volatile solute. Hence, this gives us an idea that the solution remains at 100 gms from which 2 gms (W₂) is regarded as the non-volatile solute & the desired solvent stands at 98 gms (W₁).
Then, the overall molar mass of water will be, (M₁) = 18 g/mol. Considering Raoult’s law for dilute solutions; we get,

Finally, the molar mass of the solute is defined by (M₂) = 41.34 g/mol.

1.16

Ans – Given,
Vapour Pressure of heptane = 105.2 kPa
Vapour Pressure of octane = 46.8 kPa
Mass of heptane = 26 g
Mass of octane = 35 g
Molar mass of heptane = 100 g/mol
Molar mass of octane = 114 g/mol

Vapor pressure of heptane = x_H × P⁰ = 0.458 × 105.2
= 48.1816 kPa 
Vapor pressure of octane = x_O × P⁰ = 0.541 × 46.8
= 25.3188 kPa 
∴ Vapor pressure of mixture = 48.1816 + 25.3188 = 73.5004kPa

1.17

Ans – One molal solution is one mole of solute in 1000g of solvent.

1.18

Ans – According to Raoult’s Law,

1.19

Ans – Let molar mass of solute be M g mol^-1

1.20

Ans – In 5% (by mass) solution, mass of sugar denotes 5g in 100g of solvent (water).

1.21

Ans – By using the relation we know,

The atomic masses of A and B are denoted as ‘p’ and ‘q’, respectively.

1.22

Ans –

1.23

Ans – (i) n-hexane & n-octane – These 2 are nonpolar molecules. So, the intermolecular interaction will happen in London-oriented dispersion forces.

(ii) I₂ & CCl₄ – These 2 molecules are nonpolar & the intermolecular interactions will happen as London dispersion forces.

(iii) NaClO₄ & H₂O – The intermolecular interactions are completely based on ion-dipole interactions where NaClO₄ is an ionic compound and the H₂O element is a polar molecule.

(iv) Methanol & Acetone – These 2 elements are polar molecules and their intermolecular conversion might occur as dipole-dipole interactions.

(v) Acetonitrile (CH₃CN) & Acetone (C₃H₆O) – The aforementioned elements are polar molecules and the intermolecular interactions in them occur as dipole-dipole-based interactions.

1.24

Ans – The segregation of elements based on the ascending order of solubility in n-octane can be defined as,
KCl < CH₃OH < CH₃CN < Cyclohexane.

It happens due to the below factors,

• KCl is an ionic element and won’t be dissolved in n-octane.
• CH₃OH is a polar molecule & may get dissolved in n-octane.
• CH₃CN is another polar molecule but has a lesser value when compared to CH₃OH. Hence, it might get dissolved in n-octane to a considerable range.
• Cyclohexane is also a polar & basic molecule that impacts its dissolving features across different proportions in the group.

1.25

Ans – (i) Phenol – Remains insoluble in H₂O

(ii) Toluene – It isn’t soluble with one another because of its non-polar specification.

(iii) Formic Acid – It remains partially soluble in water.

(iv) Ethylene Glycol – It is partially soluble in water.

(v) Chloroform – It remains highly insoluble in water.

(vi) Pentanol – Pentanol is partially soluble in water because of its presence of the alcohol functional group.

1.26

Ans –

1.27

Ans – Maximum molarity of CuS = Solubility of CuS
Solubility of CuS will be S mol L^-1

1.28

Ans –

1.29

Ans – As we have mentioned,
The molality of solution = 1.5 × 10⁻³ m shows us that the 1.5 × 10⁻³ moles are still available in 1 kg of solvent.
Molar mass of nalorphine = 311 g/mol
Hence, The mass of solute = 1.5 × 10⁻³ × 311 = 0.4665g
Total mass of the solution = 0.4665 + 1000 = 1000.4665 g
Based on the above equations, we can conclude that 0.4665 g Narlophene demands a total solution of 1000.4665 gms.
Therefore, 1.5 mg of Narlophene might consist of,

1.30

Ans – As we all know,
We also contain a 0.15M solution depicting the solution of 1000 ml with 0.15 moles of solute.
Hence, to build 250 ml of solution we might need = (250 × 0.15)/1000 = 0.0375mol 
The total molar mass of benzoic acid = 122 g/mol
So, the total mass of benzoic acid can be defined as,
Mass = 0.0375 × 122 = 4.575g

1.31

Ans – As we know, fluorine has been regarded as highly electronegative when compared to chlorine, which has the maximum number of electron-withdrawing inductive effects. Hence, trifluoroacetic acid will be considered as the strongest trichloroacetic acid. The second most common one, acetic acid is deemed as the weakest acid of all due its the absence of an electron-withdrawing pair of ions. So, F₃CCOOH ionizes to the maximum possible threshold whereas CH₃COOH ionizes to a minimum considerable threshold in water. The higher the value of ionization, the more massive the depression in the freezing point will get. That is the reason why the depression order in the freezing point can be framed as CH₃COOH < Cl₃CCOOH < F₃CCOOH.

1.32

Ans – As you all know,
The total mass of CH₃CH₂CHClCOOH is 10 g
Molar mass of CH₃CH₂CHClCOOH = 122.5 g/mol
No. of moles of the same will be n = 10/122.5 = 0.0816moles
The total mass of water (solvent) can be defined as W₁ = 250 g

Hence,
The total moles that remain at equilibrium = 1 – α + 2α = 1 + α
Meanwhile, the Van’t Hoff factor can be defined as i = 1 + α
Finally, placing the total values we obtain will be i = 1 + 0.06548 = 1.06548
Ultimately, the reduction in freezing point can be mentioned as,

1.33

Ans – Given the values,
Mass of water, w₁ = 500 g = 0.5 kg
Mass of fluoroacetic acid (CH₂FCOOH), w₂ = 19.5 g
Molal freezing point depression constant for water, K_f = 1.86 K kg mol^-1
ΔT_f = 1°C
Molar mass of CH₂FCOOH (M₂)
= 2 × 12 + 3 × 1 + 1 × 19 + 2 × 16
= 24 + 3 + 19 + 32 = 78 g mol^-1

Considering 500 mL as the solution’s volume, we have:

1.34

Ans – Apply Raoult’s Law

1.35

Ans – Given values are Henry’s law constant, K_H  = 4.27×10⁵ mm Hg and Pressure = 760 mm Hg.
Using Henry’s law we can calculate the molality as,

1.36

Ans – For liquid A number of moles, n_A = w₁/M₁ = 100/140 mol = 0.714 mol
For liquid A number of moles, n_B = w₂/M₂ = 1000/180 mol = 5.556 mol
Mole fraction of A will be,

Mole fraction of B will be, x_B = 1 – x_A = 1 – 0.114 = 0.886
For pure liquid B vapour pressure, P⁰_B = 500 torr
⇒ In the solution vapour pressure of liquid B will be,
P_B = P⁰_B x_B = 500 × 0.886 = 443 torr
Given P_total = 475 torr
⇒ P_A = P_total – P_B
= 475 – 443 = 32 torr
In the solution vapour pressure of liquid A will be,
P_A = P⁰_A x_A
⇒ Vapour pressure of pure liquid A, P⁰_A = P_A/x_A
= 32/0.114 = 280.7 torr

1.37

Ans – The following data can be derived from the provided table;

The graph can be represented as follows;

The table and graph indicate that the solution exhibits a negative deviation from ideal behaviour, as evidenced by the downward trend of P_total.

1.38

Ans – From the given data we get,
Benzene mass = 80 g and Molar mass = 78 g/mol
⇒ Number of moles of benzene, n_B = 80/78
= 1.025 moles
Similarly toluene mass = 100 g and Molar mass = 92 g/mol
⇒ Number of moles of benzene, n_T = 100/92
= 1.086 moles
Mole fraction of benzene,

Mole fraction of toluene, x_T = 1 – x_B = 1 – 0.4855 = 0.5144
Given vapor pressures P⁰_B = 50.71 mm Hg and P⁰_T = 32.06 mm Hg
Using raoult’s law,
P_B = P⁰_B x_B = 50.71 x 0.4855 = 24.61 mm Hg
P_T = P⁰_T x_T = 32.06 x 0.5144 = 16.491 mm Hg

1.39

Ans – % of oxygen (O₂) in air = 20%
% of nitrogen (N₂) in air = 79%
Also, it is given that water is in equilibrium with air at a total pressure of 10 atm = (10 × 760) mm = 7600 mm
∴ partial pressure of oxygen,
PO₂ = 20/100 x 7600 mm
= 1520 mm Hg
Partial pressure of nitrogen,
pN₂ = 79/100 x 7600 mm
= 6004 mm Hg
According to Henry’s law,
p = K_H.χ

1.40

Ans – Given values are
Volume, V= 2.5 L
Van’t Hoff factor, i = 2.47
Osmotic pressure, π = 0.75 atm
Gas constant, R = 0.0821 L atm K⁻¹ mol⁻¹ 
Temperature, T = 273 + 27 = 300 K
Molar mass of CaCl₂, M = 111g/mol
The formula for osmotic pressure is

1.41

Ans –

Number if ions produced = Van’t Hoff factor = i = 3
∴ Osmotic pressure is given as,

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