W₂ = 2.56 g
W₁ = 100 g
T = 319.692 K
K_b = 2.42 K Kg mol^–1
\(\Delta\)T_b = (319.692 – 319.450) K = 0.242 K
\(M_2={K_b\times W_2\times 100\over \Delta T_b\times W_1}\)
\(={2.42\times 2.56\times 1000\over 0.242\times 100}\)
M₂ = 256 g mol^-1
Molecular mass of sulphur in solution = 256 g mol^–1
atomic mass of one mole of sulphur atom = 32
No. of atoms in a molecule of sulphur = \({256\over 32}=8\)
Hence molecular formula of sulphur is S₈.

\(P_{pure\ benzene}^o=\) 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.

Total pressure = 1 atm
\(P_{N_2}=({80\over 100})\times total\ pressure={80\over 100}\times 1\ atm=0.8\ atm\)
\(P_{O_2}=({20\over 100})\times 1=0.2\ atm\)
Accordingg to Henry's Law
P_solute = K_H X_solute in solution
\(\therefore\) \(P_{N_2}=(K_H)_{nitrogen}\times \) mole fraction of Nitrogen in solution
\({0.8\over 8.5\times 10^4}=X_{N_2}\)
\(X_{N_2}=9.4\times 10^{-6}\)
Similarly,
\(X_{O_2}={0.2\over 4.6\times 10^4}\)
= 4.3 x 10^-6.

\({ P }_{ { C }_{ 6 }{ H }_{ 6 } }^{ 0 }=\) 640 mm Hg
W₂ = 2.2 g (non volabile solute)
W₁ = 40 g (benzene)
P_solution = 600 mm Hg
M₂ = ?
\({P^o-P\over P^o}=X_2\)
\({640-600\over 640}={n_2\over n_1+n_2}\)   [\(\therefore n_1>>n_2;n_1+n_2\approx n_1\)]
\({40\over 640}={n_2\over n_1}\)
\(0.0625={W_2\times M_1\over M_2\times W_1}\)
\(M_2={2.2\times 78\over 0.0625\times 40}\)
= 68.64 g mol^-1.

(k_H)_bonzene = 4.2 x 10^–5 mm Hg
Solubility of methane = ?
P = 750mm Hg           
P = 840 mm Hg
According to Henrys Law,
P = K_H . x_{in solution}.
750 mm Hg = 4.2 x 10^–5 mm Hg . x_{in solution}
\(\Rightarrow X_{insolution}={750\over 4.2\times 10^{-5}}\)
i.e, solubility = 178.5 x 10⁵
similarly at P = 840 mm Hg
solubility = \({840\over 4.2\times 10^{-5}}\)
= 200 x 10^-5.