It is the ratio of the number of moles of a component to the total number of moles of all the components present in a solution.
\(\text { Mole fraction of a components }=\frac{\text { No. of moles of the components }}{\text { Total no. of moles of all the components present in a solution } }\)
\(\mathrm{X}_{\mathrm{A}}=\frac{\mathrm{n}_{\mathrm{A}}}{\mathrm{n}_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}}}\).

When an inert gas is added to an equilibrium system at constant volume, the total number of moles of gases present in the container increases, that is, the total pressure of gases increases the partial pressure of the reactants and the products are unchanged. Hence at constant volume addition of inert gas has no effect on equilibrium.

1.
2. The compound with more number of electrons on the valence shell,  \(\therefore\)  is considered to be the most electronegative element.
\(\therefore\) HZ is the most electronegative element which has highest dipole moment.

The heat of solution is defined as "the change in enthalpy of the system when one mole of a substance is dissolved in a specified quantity of solvent at a given temperature".

The sizes of oxygen and fluorine atoms are comparatively small and they have high electron density. Moreover, the extra electron added to oxygen and fluorine has to be accommodated in the 2p orbital which is relatively compact compared to the 3p orbital of sulphur and chlorine so, oxygen. and fluorine have lower electron affinity than their respective group elements sulphur and chlorine.

The electronic configuration of sulphur is 1s² 2s² 2p⁶ 3s² 3p⁴
For 1s², n,+ l = 1+ 0 = 1 ; For 2p⁶, n + 1 = 2 + 1 = 3
2s², n + I = 2 + 0 = 2;      For 3s², n + I = 3 + 0 = 3
3p⁴, n + 1 = 3 + 1 = 4
Thus n + 1 = 3 for 2p⁶ electron and 3s² electron = 8 electrons.

The low solubility of LiF in water is due to its high lattice enthalpy (small size of Li⁺ and F^-) than its solvation (hydration) energy.

 (ii) \({ K }\overset { +7 }{ \underset { \underset { 3e^{ - } }{ \uparrow } }{ M } } n{ O }_{ 4 }+{ Na }_{ 2 }\overset { +4 }{ \underset { { 2e }^{ - } }{ \underset { \downarrow }{ S } } } { O }_{ 3 }\longrightarrow \overset { +4 }{ M } { nO }_{ 2 }+{ Na }_{ 2 }\overset { +6 }{ s } { O }_{ 4 }+KOH\)
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)   MnO₂ + Na₂SO₄ + KOH
\(\Rightarrow\) 2KMnO₄ + 3Na₂SO₃ \(\longrightarrow \)  2MnO₂ + 3Na₂SO₄ + KOH
\(\Rightarrow\) 2KMNO₄ + 3NaSO₃ + H₂O \(\longrightarrow \) 2MnO₂ + ₃Na₂SO₄ +2KOH

Units for Vanderwaal's constant:
The dimensions of the vanderwaal's constant a and b depend upon the units of P and V respectively.
\(a=\frac { Pressure.({ Volume) }^{ 2 } }{ { mole }^{ 2 } } \)
a = atm . dm⁶ . mol^-2 (1 litre = 1 dm³)
If volume is expressed in dm³. Then b is expressed as,
\(\frac { Vol }{ n } =\frac { { dm }^{ 3 } }{ mol } \)
\(\frac { V }{ n } ={ dm }^{ 3 }{ mol }^{ -1 }\)
unit of b = dm³ mol^-1

Electrolysis of heavy water gives deuterium.
\({ 2D }_{ 2 }O\overset { Electrolysis }{ \longrightarrow } { 2D }_{ 2 }+{ O }_{ 2 }\)