A solution is a homogeneous mixture of two or more I substances, consisting of atoms, ions or molecules. The compound that is present in largest amount in a homogeneous mixture is called the solvent, and the others are solutes. For example, when a small amount of NaCI is dissolved in water, a homogeneous solution is obtained. In this solution, Na+ and Cl^- ions are uniformly distributed in water. Here water is the solvent as the amount of water is more compared to the amount of NaCI present in this solution, and the NaCl,is the solute.

\(\Delta T_f=K_f\ m\)
ie \(\Delta T_f\alpha\ m\)
\(m_{CH_3-OH}={({10\over 32})\over 0.1}=3.125\ m\)
\(m_{C_2H_5-OH}={({20\over 46})\over 0.2}=2.174\ m\)
\(\therefore\) depression in freezing point is more in methanol solution and it will have lower freezing point.

ΔT_b = K_b m
= K_b x W₂ x 1000 / M₂ x W₁
M₂ = K_b x W₂ x 1000 / ΔT_b x W₁
= 7.5 x 0.75 x 1000 / 0.15 x 200
= 187.5 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.

Let the volume of 4M HCl required to prepare 500 mL of 2.5 MHCl = x mL
Therefore, the required volume of 2M HCl = (500 - x) mL
We know from the equation
C₁V₁+ C₂V₂ = C₃V₃
(4x) + 2(500 - x) = 2.5 x 500
4x + 1000 - 2x = 1250
2x = 1250 - 1000
\(x={250\over 2}\)
= 125 ml
Hence, volume of 4M HCl required = 125 mL
Volume of 2M HCl required = (500 - 125) mL = 375 mL.

\(10\%{W\over V}\) means that 10g of solute in 100 ml solution
\(\therefore\) amount of iodopovidone in 1.5 ml = \({10\ g\over 100\ ml}\times 1.5\ ml\)
= 0.15 g.

Solution is a homogeneous mixture of two or more substances. If the water is the solvent in the solution, then it is an aqueous solution.

Generally, liquids have a tendency to evaporate. If the kinetic energy of molecules in the liquid state overcomes the intermolecular force of attraction between them, then the molecules will escape from the liquid state. This process is called 'evaporation' and it happens on the surface of the liquid.

It is the ratio of number of parts of the components to the total number of parts of all the components per million.

Osmosis, which is a spontaneous process by which the solvent molecules pass through a semi permeable membrane from a solution of lower concentration to a solution of higher concentration.

Molality : It is the number of moles of the solute present one kg of the solvent
Molality = \(\frac { No.of \ moles\ of \ solute }{ Mass\ of\ the\ solvent\ (in\ kg) } \) 

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 = \({(\Delta T_f)\ observed\over (\Delta T_f)\ calculated}={0.680\ K\over 0.372\ K}=1.82\)

(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.

Generally the change in pressure does not have any significant effect in the solubility of solids and liquids as they are not compressible. However, the solubility of gases generally increases with increase of pressure.
Consider a saturated solution of a gaseous solute dissolved in a liquid solvent in a closed container. In such a system, the following equilibrium exists.
Gas (in gaseous state) = Gas (in solution)
According to Le-Chatelier principle, the increase in pressure will shift the equilibrium in the direction which will reduce the p.ressure. Therefore, more number of gaseous molecules dissolves in the solvent and the solubility increases.

Henry's law states that, "the partial pressure of the gas in vapour phase is directly proportional to the mole fraction(x) of the gaseous solute in the solution at low concentrations".
Henry's law can be expressed as,
\(\rho \)_solute \(\alpha\) X _{solute in solution}
P_solute = K_Hx _{solute in solution}
Explanation: Here, P_solute represents the partial pressure of the gas in vapour state which is commonly called as vapour pressure. x _solute in solution represents the mole fraction of solute in the solution. K_H is a empirical consiint with the dimensions of pressure. The value of 'K_H' depends on the nature of the gaseous solute and solvent. The above equation is a straight-line in the form of y = mx. The plot partial pressure of the gas against its mole fraction in a solution will give a straight line as shown in fig The slope of the. straight line gives the value of K_H.

Limitation of Henry's law:
i) Henry's law is applicable at moderate temperature and pressure only.
ii) Only the less soluble gases obeys Henry's law.
iii) The gases reacting with the solvent do not obey Henlry s law For example, ammonia or HCI reacts \Mith water and hence does not obey this law.
\(\mathrm{NH}_3+\mathrm{H}_2 \mathrm{O} \leftrightarrows \mathrm{NH}_4^{+}+\mathrm{OH}^{-}\)
(iv) The gases obeying Henry's law should not associate or dissociate while dissolving in the solvent.