Class 12 Chemistry Chapter 1 Solutions – NCERT Answers & Key Concepts

Given-

Mass of K₂SO₄, w = 25 mg = 25 X 10^-3 g,

Molar mass of K₂SO₄ = (39×2) + (32×1) + (16×4) = 174 g mol^-1

Volume V = 2 liter

T = 25⁰C + 273 = 298 K (add 273 to convert in Kelvin)

The reaction of dissociation of K₂SO₄ is written as,

K₂SO₄ → 2K ⁺ + SO₄^2-

Number if ions produced = 2 + 1 = 3, hence vant Hoff’s factor, I = 3

Here, we use vant Hoff’s equation for dilute solutions, given as,

^{πV = inRT}

where, n is the number of moles of solute, R is solution constant which is equal to the gas constant (0.082) and T is the absolute temperature (298 K).

Hence, the osmotic pressure of a solution is 5.27x10^-3atm

Given,

Volume of water, V = 450 mL = 0.45 L

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

1.0 g of polymer of molar mass 185,000

Number of moles of polymer, n = 1 / 185,000 mol

We know that,

Osmotic pressure? = nRT/V

= 1 X 8.314 X 10³ X 310 / 185000 X 0.45

= 30.98 Pa

= 31 Pa (approx)

Given: mass of solute = 30g

Let the molar mass of solute be x g and vapour pressure of pure water at 298k be P₁ ^?

Mass of water(solvent) = 90g

Molar mass of water = H₂O = 1 × 2 + 16 = 18g

Moles of water = mass of water/molar mass

⇒ n = 90/18 moles

⇒ n = 5moles

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

x₂ = (30/x) / (30/x) + 5

x₂ = 30 / 30+5x

Vapour pressure of solution (p₁) = 2.8kpa

Applying the formula:

According to second condition when we add 18g of water to solution vapour pressure becomes 2.9kpa

Moles of water = mass/molar mass

⇒ n = 90 + 18/18

⇒ n = 6moles

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

x₂ = (30/x) / (30/x) + 6

x₂ = 30 / 30+6x

Applying the formula:

Solving 1 and 2:

Dividing (2) by (1) we get

Substituting value of x in 1 we get

(i) molar mass of the solute = 78g

(ii) vapour pressure of water at 298 K = 537kpa

Given

Mass of acetic acid, w₁ = 75 g

Lowering of melting point? T_f = 1.5 K

K_f = 3.9 K kg/mol

Molar mass of ascorbic acid (C₆H₈O₆), M₂ 6 * 12 + 8 * 1 + 6 * 16 = 176 g/mol

We know that,

= 5.08 g

Hence,

5.08 g of ascorbic acid is needed to be dissolved.

Mass of ions = 92g

Molar mass of ions = Na⁺ = 23g (neglect the mass lost due to absence of a electron)

Moles of ions = mass of ions/molar mass

⇒ n = 92/23 moles

⇒ n = 4moles

Molality of solution = moles of solute/mass of solvent (in kg) Molality = 4/1 = 4M

Given,

Vapour pressure of water, P_I^o = 23.8 mm of Hg

Weight of water, w₁ = 850 g

Weight of urea, w₂ = 50 g

Molecular weight of water, M₁ = 18 g/mol

Molecular weight of urea, M₂ = 60 g/mol

n₁ = w₁/M₁ = 850/18 = 47.22 mol

n₂ = w₂/M₂ = 50/60  = 0.83 mol

We have to calculate vapour pressure of water in the solution p₁

By using Raoult's therom,

P_I = 23.4 mm of Hg Hence,

The vapour pressure of water in the solution is 23.4mm of Hg and its relative lowering is 0.0173.

Molarity = Moles of Solute / Volume of Solution in liter

(a) Given, In 4.3 L of solution there is 30 g of Co (NO₃)₂. 6H₂O

Molar mass of Co (NO₃)₂.6H₂O = (1 × 59 + 2 × (1 × 14 + 3 × 16) + 6 × 18)

= 291 g/mol.

∴ Moles = Given Mass / Molar Mass = 30/291 = 0.103 mol.

Now, Molarity = 0.103 mol / 4.3 L

= 0.023 M

(b) Given, 30 mL of 0.5 M H₂SO₄ diluted to500 mL.

In 1000 mL of 0.5 M H₂SO₄, number of moles present is 0.5 mol.

∴ In 30 mL of 0.5 M H₂SO₄, number of moles present = 30X 0.5 / 1000 mol.

= 0.015 mol.

∴ Molarity = 0.015 mol / 0.5L

= 0.03 M.

(i) Both the compounds are non-polar and they do not attract each other because they do not form any polar ions. Vanderwaals forces of attraction will be dominant in between them as vanderwaals forces of attraction are not a result of any chemical or electronic bond.

(ii) now here both the compounds are non-polar because in I₂ both the atoms are same so they have same electronegativity and hence there will be no displacement of electron cloud, it will be in the centre. In case of CCl₄ molecule, it has tetrahedral shape so two Cl atoms will cancel the attraction effect from two opposite Cl atoms, hence molecule as a whole is non polar. Therefore they will also have vanderwaals forces of attraction.

(iii) NaClO₄ is ionic in nature as Na, Cl and O all have different electronegativity and their shape is also not symmetric. So molecular will be ionic in nature and we know that water is polar because O will attract the electron cloud towards it (more electronegative) hence there will be formation of dipole (two oppositely charged ions separated by a short distance). Therefore there will be ion-dipole interaction between them.

(iv) methanol and acetone are both polar molecules because of the presence of electron withdrawing O atom in methanol and ketone group in acetone. So they will have dipole-dipole interaction.

(v) acetonitrile is polar compound due to presence of electronegative N atom and acetone due to ketone group. So, dipole-dipole interaction.

Now n-octane is non-polar solvent due to long chain saturated structure. We know that “like dissolves like” so a non-polar compound will be more soluble in non-polar solvent as compared to polar compound.

So cyclohexane is non-polar due to symmetric structure. KCl is ionic in nature as it will dissociate into K ⁺ and Cl^- ions. CH₃CN is polar as mentioned above and CH₃OH is also polar in nature.

The order of increasing polarity is:

Cyclohexane < CH₃CN < CH₃OH < KCl (O is more electronegative than N)

Therefore, the order of increasing solubility is:

KCl < CH₃OH < CH₃CN < Cyclohexane

According to Henry's law,"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."
Stated as,  
p = K_Hx 
Where, P = partial pressure of the solute above the solution
K_H = Henry's constant
x = concentration of the solute in the solution
Given,
Solubility of H₂S in water at STP is 0.195 m
We know,
At STP pressure p = 0.987 bar
0.195 mol of H₂S is dissolved in 1000g of water

Moles of water = 1000/18
 = 55.56 g/mol
∴ the mole fraction of H₂S = Moles of H₂S / Moles of H₂S + Moles of water
 = 0.195 / 0.195+55.6
 = 0.0035
According to Henry's law, p = K_Hx 
K_H = p/x
K_H = 0.987 / 0.0035
K_H = 282 bar 
∴ The Henry’s law constant is 282 bar

According to Henry's law, "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."
Stated as,
p = K_Hx 
Where,  
P = partial pressure of the solute above the solution
K_H = Henry's constant
x = concentration of the solute in the solution
Given,

K_H = 1.67x10⁸ Pa 

P_CO2  2.5 atm = 2.5 * 1.01325 * 10⁵Pa

According to Henry's law,

p = K_Hx

x = P/K_H

x = 2.5 X 1.01325 X 10⁵ / 1.67 X 10⁸

x = 0.00152

In 500 ml of soda water there is 500 ml of water (neglecting soda)

Mole of water = 500/18

= 27.78 mol

Now,

x = n_co2 / n_co2 + n_H2o

n_co2  = x n_H2o

 = 0.00152 X 27.78

= 0.042 mol

Hence, quantity of CO₂ in 500mL of soda water  0.042 * 44 = 1.848 g

2.5 kg of 0.25 molal aqueous solution.
Molar mass of urea (NH2CONH2) = (2 (1 * 14 + 2 * 1) + 1 * 12 + 1 * 16)
 = 60 g/mol
1000 g of water contains 0.25 mol = (0.25 * 60) g of urea.
 = 15 g of urea.
Means, 1015 g of solution contains 15 g of urea

Therefore,
2500 g of solution contains = 15 X 2500 / 1015
 = 36.95 g
Hence, mass of urea required is 37 g (approx).

Given, P_A^o = 450 mm Hg

^P_Bo = 700 mm Hg

p_total = 600 mm of Hg

By using Rault's law,

p_total = P_A + P_B

p_total = P_A^ox_A + P_B^ox_B

p_total = P_A^ox_A + P_B^o ( 1 - xA )

p_{total = (}P_A^o- P_B^o)x_{A +} P_B^o

600 = (450 - 700) x_A + 700

-100 = -250 x_A

x_A = 0.4

∴ x_B = 1 - x_A

x_B = 1 – 0.4

x_B = 0.6

Now,

P_A = P_A^ox_A

P_A = 450 × 0.4

P_A = 180 mm of Hg and

P_B = P_B^ox

P_B = 700 × 0.6

P_B = 420 mm of Hg

Composition in vapour phase is calculated by

Mole fraction of liquid,

A =P_A / P_A + P_B

= 180/180+420

= 0.30

Mole fraction of liquid,

B =P_B / P_A + P_B

= 420 / 180+420

= 0.70

5% solution means 5g of cane sugar is present in 100g of solution

Freezing point of solution = 271k

Freezing point of pure water = 273.15k

Molar mass of cane sugar (C₁₂H₂₂O₁₁) = 12 × 12 + 1 × 22 + 16 × 11 = 342g

Moles of cane sugar = mass/molar mass = 5/342

⇒ n = 0.0146mol

Molality of solution = moles of solute/mass of solvent (in kg)

⇒ M = 0.0146/0.095

⇒ Molality = 0.154M

Depression in freezing point = ΔT_f = 273.15-271 = 2.15k

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ K_f = 2.15/0.154

⇒ K_f = 13.96k kg mol^-1

Second condition: mass of glucose = 5g

Molar mass of glucose (C₆H₁₂O₆) = 12 × 6 + 1 × 12 + 16 × 6 = 180g

Moles of glucose = mass/molar mass

⇒ n = 5/180moles

⇒ n = 0.0278mol

Molality of solution = moles of solute/mass of solvent (in kg)

⇒ molality = 0.0278/0.095

⇒ M = 0.2926M

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ ΔT_f = 13.96 × 0.2926

⇒ ΔT_f = 4.08k

Freezing point of solution = 273.15 + 4.08 = 277.234k

Given- w₁ = 500g

W₂ = 19.5g

K_f = 1.86 K kg mol^-1

Molar mass of CH₂FCOOH = 12 + 2 + 19 + 12 + 16 + 16 + 1

= 78 g mol^-1

The depression in freezing point is calculated by,

→ (where, m is the molality)

= 1.86 X 19.5 / 78 X 1000/500

= 1.86 X 19.5 / 78 X 2

=0.93

∴ Δt_f (calculated) = 0.93

To find out the vant Hoff’s factor, we use the formula,

i = observed Δt_f / calculated Δt_f

_{i = 1.0 (given) / 0.93}

∴ i= 1.07

CH₂FCOOH → CH₂FCOO^- + H ⁺

To find out the degree of dissociation α, we use

Thus, the vant Hoff’s factor is 1.07 an the dissociation constant is 2.634x10^-3

Given- Vapour pressure of water,

P_A⁰  = 17.535 mm Hg  

W_B= 25 g of glucose

W_A = 450g of water

Molar mass of water, H₂O = 1 + 1 + 16 = 18 g mol^-1

Molar mass of glucose, C₆H₁₂O₆ = (12*6) + (1*12) + (16*6) = 180 g mol^-1

Using Raoult's law for solution of non-volatile solute,

P_A⁰ - P_A / P_A⁰ = x_B? Equation 1

where x_B is the mole fraction of the solute

x_B = W_B/M_B X M_B/W_B

=25/180 X 18 / 450

x_B = 1/180

Substituting the value of x_B in equation 1, we get,

Thus, the vapour pressure of water at 293 K at the given conditions is 17.437 mm Hg

Here,

T = 300 K

π = 1.52 bar

R = 0.083 bar L

Applying the relation, π = CRT

where

π = osmotic pressure of solution

C = concentration of solution

R = universal gas constant

T = temperature

⇒C  = π / RT = 1.52 / 0.083 X 300

⇒ C = 0.061mol/L

Concentration of the solution is 0.061mol/L

The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the pattern,

Acetic acid< trichloroacetic acid< trifluoroacetic acid.

This is because fluorine is more electronegative than chlorine. So, trifluoracetic acid is a stronger acid in comparison to trichloroacetic acid and acetic acid. And also, acetic acid is the weakest of all.

Explanation: Stronger acid produces more number of ions, therefore it has more? T_f (depression in freezing point), hence lower freezing point. As the acidic strength increases, the acid gets more and more ionised.

Trifluoracetic acid ionizes to the largest extent. Hence, in this case, trifluoracetic acid being the strongest acid produces more number of ions (extent of ionisation and concentration of ions are more), high? T_f (depression in freezing point)and lower freezing point and vice versa.

Let the total mass of the solution be 100 g and the mass of benzene be 30 g.

∴ Mass of carbon tetrachloride = (100 - 30) g = 70 g

Molar mass of benzene (C₆H₆) = (6 × 12 + 6 × 1) g mol ^-1

= 78 g mol ^-1

∴ Number of moles of C₆H₆ =30/78 mol

= 0.3846 mol

Molar mass of carbon tetrachloride (CCl₄) = 1 × 12 + 4 × 35.5

= 154 g mol ^-1

∴ Number of moles of CCl₄ = 70/154 mol

= 0.4545 mol

Thus, the mole fraction of C₆H₆ is given as:

Number of moles of C₆H₆ / Number of moles of C₆H₆  + Number of moles of CCl₄

= 0.3846 / (0.3846 + 0.4545)

Total mass of solution = 6.5g + 450g = 456.5g

Therefore mass percentage of aspirin in solution = (mass of aspirine/total mass of solution) = 6.5/456.5

⇒ mass % of aspirine = 1.424%

Given-

Vant Hoff’s factor, I = 2.47

osmotic pressure, π = 0.75 atm

Volume of solution = 2.5L.

To determine the amount of CaCl₂, we use vant Hoff’s equation for dilute solutions, given as,

πV = inRT

where, n is the number of moles of solute, R is solution constant which is equal to the gas constant and T is the absolute temperature.

Hence, the amount of CaCl₂ dissolved is 3.425g

Mass of Solution = Mass of Benzene + Mass of Carbon Tetrachloride

= 22 g + 122 g = 144 g

Mass percentage of Benzene = Mass of Benzene / Mass of Solution X 100 = 22/144 X 100 = 15.28%

Mass percentage of CCl₄ = Mass of CCl₄ / Mass of Solution X 100 = 122/144 X 100 = 84.72%

The P_total for the values given in the graph is found out and plotted in the graph.

It can be observed from the graph that the plot for the p total of the solution curves downwards. Therefore, the solution shows negative deviation from the ideal behaviour.

Given-

Mass of liquid A, W_A = 100g, Molar mass, M_A = 140 g mol^-1

Mass of liquid B, W_B = 1000 g, Molar mass, M_B = 180 g mol^-1

Using the formula below calculate the no. of moles in liquid A and B.

Number of moles = Mass / Molar Mass

Number of moles of liquid A, M_A = 100/140 = 0.714 mol^-1

Number of moles of liquid B, M_B = 1000/ 180 = 5.556 mol^-1

Using the formula,

mole fraction of a liquid = No. of moles of the liquid / total no of moles

we calculate the mole fraction of liquids A and B.

→ Mole fraction of A,

x_A = 0.714 / (0.714 + 5.556)

∴ x_A = 0.114

→ Mole fraction of B,

x_B = 1- x_{A = 1 - 0.114}

∴ x_B = 0.886

Vapour pressure of pure liquid B, P^o = 500 torr (given)

According to Henry's law,

Having given the total vapour pressure of the solution, P_total = 475 torr,

Thus, the vapour pressure of pure liquid A = 280.7 torr and

vapour pressure of liquid A in the solution = 32 torr

(a) Molality, also called molal concentration, is a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent. Molar mass of KI = 39 + 127 = 166 g/mol.
20% aqueous solution of KI means 200 g of KI is present in 1000 g of solution. Therefore,  

Molality = Moles of KI / Mass of Water in kg

= (200/166) / (0.8) = 1.506 m

(b) Molarity is the concentration of a solution expressed as the number of moles of 
solute per litre of solution.
Given,
Density of the solution = 1.202 g/mL
Volume of 100 g solution = mass/ density
 = 100/1.202 
 = 83.19 mL
Therefore, molarity = 20/166 mol/ 83.19 × 10^-3 L 
 = 1.45 M
∴ Molarity of KI = 1.45M

(c) Molar mass of KI = 39 + 127 = 166 g/mol.
Moles of KI = 20/166 = 0.12 mol
Moles of water = 80/18 = 4.44 mol

Therefore,
Mole fraction of KI = Moles of KI / Moles of KI+ Moles of Water
 = 0.12 / 0.12+ 4.44
 = 0.0263
∴ Mole fraction of KI = 0.0263

Given:

Molar mass of non-volatile solute = 40g

Let no. of moles of solute be n.

Mass of octane = 114g

Molar mass of octane (C₈H₁₈) = 12 × 8 + 1 × 18 = 114g/mol

Moles of octane = given mass/molar mass

⇒ n = 114/114 moles

⇒ n = 1 mole

Molar fraction of solute,

x₂ = moles of solute / moles of solute + moles of octane

⇒ x₂ = n/n + 1

Let the vapour pressure of original solvent (without solute) be p₁^?

Accordingly after addition of solute vapour pressure of solution reduces to 80% i.e.

0.8 p₁^? = p₁

Applying the formula:

⇒ n/n + 1 = 0.2

⇒ 0.2n + 0.2 = n

⇒ n = 0.25 moles

Hence, mass of solute is:

moles = given mass/molar mass

⇒ 0.25moles = mass/40g

⇒ mass = 10g.

Given-

K_H for O₂ = 3.30 * 10⁷ mm Hg,

K_H for N₂ = 6.51 * 10⁷ mm Hg

Percentage of oxygen (O₂) = 20 %

Percentage of nitrogen (N₂) = 79%

Total pressure = 10 atm

Using Henry's law,

where, p is the partial pressure of gas in the solution and K_H is Henry's constant.

Thus, the mole fraction of oxygen in solution, x_oxy = 4.61x10^-5

and the mole fraction of nitrogen in solution, x_nit is 9.22x10^-5

Mass of CH₃CH₂CHClCOOH = 10 g

Mass of water = 250g

K_a = 1.4 × 10^–3,

K_f = 1.86 K kg mol^–1

Molar mass of CH₃CH₂CHClCOOH = 12 + 3 + 12 + 2 + 12 + 1 + 35.5 + 2 + 16 + 16 + 1

= 122.5 g mol^–1

Number of moles of solute = Mass of Solute / Molar Mass

→ No. of moles = 10g / 122.5 g/mol

∴ No. of moles = 8.6 X 10^–2 mol

Now, Molality is given as,

M = Number of moles of solute / kg of solvent

M= 8.6 X 10^–2 X 1000 g/mol / 250 g

M = 0.3264 kg/mol

CH₃CH₂CHClCOOH = CH₃CH₂CHClCOO^- + H ⁺

Total moles at equilibrium = (1-α) + 2 α

= 1 + α

In order to find out the depression in freezing point,  

values of I (vant Hoff’s factor) and α (degree of dissociation) are to be found out.

To find out degree of dissociation, _α

∴ I = 1.0654

Now, to find out the depression in freezing point,

Given-

Thus, mole fraction of benzene in vapour phase is 0.6744

Given,

Mass of water, w_l = 500 g

Boiling point of water = 99.63°C (at 750 mm Hg).

Molal elevation constant, K_b = 0.52 K kg/mol

Molar mass of sucrose (C₁₂H₂₂O₁₁), M₂   (11 × 12 + 22 × 1 + 11 × 16) = 342 g/mol

Elevation of boiling point ΔT_b = (100 + 273) - (99.63 + 273) = 0.37 K

We know that,

ΔT_b = K_b X 1000 X W₂ / M₂ X W₁

0.37 = 0.52 X 1000 X W₂ / 340 X 500

w₂ = 0.37 X 342 X 500 / 0.52 X 1000

w₂ = 121.67 g

Hence,

121.67 g (approx) Sucrose is added to 500g of water so that it boils at 100°C.

let the molar masses of AB₂ and AB₄ be x and y respectively.

Molar mass of benzene (C₆H₆) = 12 × 6 + 1 × 6 = 78 g/mol

Moles of benzene = mass/molar mass = 20/78

n = 0.256mol

⇒ ΔT_f = 2.3 K

K_f = 5.1K kg mol^-1

For AB₂

Applying the formula: ΔT_f = K_f × M

Where

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 2.3 = 5.1 × M₁

⇒ M₁ = 0.451mol/kg

For AB₄

Applying the formula: ΔT_f = K_f × M

ΔT_f = depression in freezing point

K_f = molal depression constant

M = molality of solution

⇒ 1.3 = 5.1 × M₂

⇒ M₂ = 0.255mol/kg

M₁ = moles of solute/mass of solvent (in kg)

M₁ = 1/x / 0.02 = 1 / 0.02x = 0.451

⇒ X = 110.86g

M₂ = moles of solute/mass of solvent (in kg)

M₂ = 1/y / 0.02 = 1 / 0.02y = 0.255

⇒ y = 196.1g

atomic mass of A be a and that of B be b g respectively.

So, AB₂ : a + 2b = 110.86

AB_4: a + 4b = 196.1

⇒ a = 25.59g

⇒ b = 42.64g

The Solubility product of CuS (ksp) = 6 × 10^-16

CuS → Cu ⁺⁺ + S^2-

Let the s be solubility of CuS in mol/L

Ksp = [ Cu ⁺⁺ ] [S²]

Ksp = solubility product

6 × 10^-16 = s × s = s²

⇒ S = 2.45 × 10^-8 mol/L

Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10^-8 mol/L

Given-

Henry's law constant K_H = 4.27X 10⁵ mm Hg,

p = 760mm Hg,

Using Henry's law,

Using the formula of lowering vapour pressure,

Thus, the solubility of methane in benzene is 0.023 moles

Volume of the solution = 250mL = 0.25L

Let the no. of moles of solute be n

Molarity = No. of moles of solute/volume of solution

⇒ 0.15 = n/0.25

⇒ n = 0.0375moles

Molar mass of C₆H₅OH = 6×12 + 5×1 + 16 + 1 = 94g

Moles = mass/molar mass

⇒ 0.0375 = m/94

Mass of benzoic acid required = 3.525g.

1. Water is a polar compound (due to electronegativity difference between O and H) . We know that “like dissolves like”. So, a non-polar compound will be more soluble in non-polar solvent as compared to polar compound.
2. Phenol has the polar group -OH and non-polar group –C₆H₅ and it can not form H bonding with water (presence of bulky non-polar group) . Thus, phenol is partially soluble in water
3. Toluene has no polar Thus, toluene is insoluble in water.
4. Formic acid (HCOOH) has the polar group -OH and can form H-bond with water. Thus, formic acid is highly soluble in water
5. Ethylene glycol (OH-CH₂-CH_2-OH) has polar -OH group and can form H-bond with Thus, it is highly soluble in water.
6. Chloroform is partly soluble as CHCl₃ is polar in nature due to high electronegativity of Cl atoms, there will be production of partial + charge on H atom so it can form H bonding with water but it is also surrounded by 3 Cl atoms, so partly soluble
7. Pentanol (C₅H₁₁OH) has polar -OH group, but it also contains a very bulky non- polar group –C₅H₁₁. Thus, pentanol is partially soluble in water

Molar mass of Nalorphene = 311g/mol

Now 1000g of solution contains 1.5 × 10^-3 moles of Nalorphene (Molality of solution = moles of solute/mass of solvent (in kg)

⇒ 1.5 × 10^-3 moles of Nalorphene = 1.5 × 10^-3 × 311 = 0.4665g of Nalorphene

Therefore, total mass of the solution = (1000 + 0.4665) g

⇒ total mass = 1000.4665 g

This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665 g.

Therefore, mass of the solution containing 1.5 mg of nalorphene is:

Mass = 1000.4665 X 1.5 X 10^-3 / 4.665 g

⇒ mass of solution containing required ions = 3.22 g

Hence, the mass of aqueous solution required is 3.22 g.

Shiksha focuses on completely fulfilling the needs of CBSE class 12 board students. We provide NCERT Solutions for classes 11th and 12th in class for physics, chemistry and mathematics.

Here are the key reasons why you should use Shiksha's NCERT Solutions for the Solution Chapter: 

• Complete coverage of NCERT Exercise, including MCQs, very short, short and long answer type questions.
• Step-by-step solutions according to the CBSE board exam needs.
• Intext question and answer with a free downloadable PDF
• complete study material for the chapter
  • NCERT Chapter 1 Solutions Notes 
  • Solutions Quick Revision Notes
  • Important Questions for Solution Chapter
  • Chemistry Chapter 1 Solutions NCERT Exemplar Solutions

NCERT offers the foundation for every class 12 student. CBSE board exam questions are primarily based on the NCERT exercise, Examples and conceptual explanation.

There are various benefits of using NCERT Solutions. 

• Specially for CBSE Board students, these NCERT Solutions are the best resource to prepare for the boards.
• These NCERT Solutions include step-wise descriptive answer approach that will help you write good answers.
• These solutions also help you understand problem-solving and how to approach a specific type of conceptual or numerical problem.
• If you are preparing for the JEE or NEEt exam, these solutions can be your first decisive step towards success.
• For NEET and JEE aspirants, NCERT solutions are the first foundation block that offer understanding of fundamental concepts, their application

After the 2023-24 revised CBSE syllabus, many chapters have been removed, and in some chapters syllabus is reduced. As per the latest CBSE exam pattern 2025-26, the Chapter 1 Solutions will be asked for 5-7 marks in the Class 12 Chemistry CBSE board exams.

For other competitive exams, the Solutions chapter has a medium weightage in the NEET exam. In the NEET exam, 1-2 questions are frequently asked based on the concept of moderate-level numerical.

Physical Chemistry holds over 30% weightage in the chemistry section in the JEE Mains exam. The solutions chapter is considered a part of physical chemistry.  You can expect 1 question form the solutions chapter in JEE Mains exam.