Solutions

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

(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.

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

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: 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