Weight of solute (W₂) = 20 mass percent of solution means 20 g of ethylene glycol
Weight of solvent (water) W₁ = 100 - 20 = 80 g
ΔT_f = K_f m
\(={K_f\times W_2\times 1000\over M_2\times W_1}\)
\(={1.86\times 20\times 1000\over 62\times 80}\)
= 7.5 K
The temperature at which the ice will begin to separate is the freezing of water after the addition of solute i.e 7.5 K lower than the normal freezing point of water (273 - 7.5K) = 265.5 K

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.

\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
In a 2 % solution weight of the solute is 2g and solvent is 98g
ΔP = P_A - P_solution = 1.013 - 1.004 bar = 0.009 bar
\(M_B={P_A^o\times W_B\times M_A\over \Delta P\times W_A}\)
M_B = 2 x 18 x 1.013/(98 x 0.009)
= 41.3 g mol^-1.

\(P_{pure\ benzene}^o=\) 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.

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.

Osmotic pressure of urea solution (\(\pi_1\)) = CRT
\(={W_2\over M_2V}RT\)
\(={6\over 60\times 1}\times RT\)
Osmotic pressure of glucose solution \((\pi_2)={W_2\over 180\times 1}\times RT\) For isotonic solution,
\(\pi_1=\pi_2\)
\({6\over 60}RT={W_2\over 180}RT\)
\(\Rightarrow W_2={6\over 60}\times 180\)
\(W_2=18\ g\)

\({ P }_{ { C }_{ 6 }{ H }_{ 6 } }^{ 0 }=\) 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.

\(\Delta T_f=0.093^oC=0.093K\)
m = ?
K_f = 1.86K Kg mol^-1
\(\Delta T_f=K_f.m\)
\(\therefore m={\Delta T_f\over K_f}\)
\(={0.093K\over 1.86\ K\ Kg\ mol^{-1}}\)
= 0.05 mol Kg^-1
= 0.05 m.

(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.