In 10^–4 M K₂SO₄ solution, there are 10^–4 moles of potassium sulphate.
K₂SO₄ molecule contains 3 ions (2K⁺ and 1So₄^2–)
1 mole of K₂SO₄ contains 3 x 6.023 x 10²³ ions
10^–4 mole of K₂SO ₄ contains 3 x 6.023 x 1023 x 10^–4 ions = 18.069 x 10¹⁹.

Molar mass of solute
\(={1000\times K_f\times mass\ of\ NaCl\over \Delta T_f\times mass\ of\ solvent}\)
\(={1000\times 1.86\times 1\over 0.24\times 200}\)
= 38.75 g mol^-1
= 38.75 g mol
Theoretical molar mass of NaCl is 
i = \({Theoretical\ molar\ mass\over Experimental\ molar\ mass}={58.5\over 38.75}=1.50\)

\(P_{pure\ benzene}^o=\) 50.71 mm Hg
\(P_{napthalene}^o=\) 32.06 mm Hg
Number of moles of benzene = \({39\over 78}\) = 0.5 mol
Number of moles of napthalene = \({128\over 128}\) = 1 mol
mole fraction of benzene = \({0.5\over 1.5}\) = 0.33
mole fraction of napthalene = 1 – 0.33 = 0.67
Partial vapour pressure of benzene = \(P_{ benzene}^o\times\) mole fraction of benzene
= 50.71 x 0.33
= 16.73 mm Hg
Partial vapour pressure of napthalene = 32.06 x 0.67 = 21.48 mm Hg
Mole fraction of benzene in vapour phase = \({16.73\over 16.73+21.48}={16.73\over 38.21}=0.44\)
Mole fraction of napthalene in vapour phase = 1 – 0.44 = 0.56.

The amount of dissolved oxygen in water decreases with rise in water's temperature. Cold water has more dissolved oxygen per unit area than warm water. So they are more comfortable is cold water during winter. During summer the warm water contain less dissolved oxygen.

P_solute  = K_H X_{solute in solution}
At pressure 1.5 atm,
p₁ = K_H x_{1       ------}(1)
At pressure 6.0 atm,
p₂ = K_H x₂        -----(2)
Dividing equation (1) by (2)
From equation p₁/p₂ = x₁/x₂
1.5/6.0 = 0.24/x₂
Therefore x₂ = 0.24 x 6.0/1.5 = 0.96 g/L.

Given: M = 12 M;
d = 1.2 gL^-1
In 12 M - HCI means 12 mole of HCI in 1 litre of the solution (i.e)
n = 12 mole
Calculation of mass of 1L of HCI solution:
Mass of 1 L of HCl solution = d x v
= 1.2 x 1000 = 1200 g
Mass of HCI (m) = No. of moles of HCl x Molar mass of HCl
= nm
= 12 x 36.5
= 438 g
.'. Mass of water (solvent) = 1200 - 438
= 762 g = 762 x 10^-3 kg
\(\therefore \text { Molality }=\frac{\text { No.of moles of solute }}{\text { Mass of solvent }(\mathrm{kg})}\)
\(=\frac{12}{762 \times 10^{-3}}\)
= 15.75 m.

i = \({observed\ property\over Theoritical\ property\ (calculated)}\)
Given ΔT_f = 0.680 K
m = 0.2 m
ΔT_f (observed) = 0.680 K
ΔT_f (calculated) = K_f m
= 1.86 K Kg mol^–1 × 0.2 mol Kg^–1
= 0.372 K
i = \({(\Delta T_f)\ observed\over (\Delta T_f)\ calculated}={0.680\ K\over 0.372\ K}=1.82\)