Ncert Solutions Chemistry Class 12th

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