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.

Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.

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

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,
\(\pi_1=\pi_2\)
\({6\over 60}RT={W_2\over 180}RT\)
\(\Rightarrow W_2={6\over 60}\times 180\)
\(W_2=18\ g\)

Molar mass = \({mass\ of\ unknown\ solute × RT\over osmotic\ pressure\ ×\ volume\ of\ solution}\)
\(= {1.5\times 8.314\times 10^{-2}\times 400\over 0.3\times 1.5}\)
= 110.85 gram mol^-1.

No. of, moles of glucose ; n₂ = \(\frac{\mathrm{m}}{\mathrm{M}}=\frac{2.82}{180}=0.016 \mathrm{~mol} \)
No. of. moles of water ; n₁ = \(\frac{30}{18}=1.67 \mathrm{~mol} \)
\(\mathrm{X}_{1}=\frac{\mathrm{n}_{1}}{\mathrm{n}_{1}+\mathrm{n}_{2}}=\frac{1.67}{1.67+0.016}=0.99\)
\(\mathrm{X}_{1}+\mathrm{X}_{2}=1 \)
\(\therefore \mathrm{X}_{2}=1-\mathrm{X}_{1}=1-0.99=0.01 \).

No.of moles ; n = \(\frac{\mathrm{m}}{\mathrm{M}}=\frac{5.6}{56}=0.1 \mathrm{~mol}\)
(i) V = 500 ml = \(\frac{500}{1000}=0.5 \mathrm{~L} \)
Molarity = \(\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.1}{0.5}=0.2 \mathrm{M} \)
(ii) V = IL 
Molarity = \(\frac{\mathrm{n}}{\mathrm{v}}=\frac{0.1}{1}=0.1 \mathrm{M} \)

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.

\(\Delta T_f=K_f\ m\)
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.

Molar mass of glycine = 1(N) + 5(H) + 2(C) + 2(O)
= 1(14) + 5(1) + 2(12) + 2(16)
= 14 + 5 + 24 + 32
= 75 g/mol
m = 7.5 g
Number of moles of glycine \( =\frac{\mathrm{m}}{\mathrm{M}} \)
\(=\frac{7.5}{75}=0.1 \mathrm{~mol}\)
Mass of solvent H₂O = 500 g
= 0.5 kg
molality = \({no.\ of\ moles\ of\ solute\over mass\ of\ solvent\ (in\ Kg)}\)
no. of moles of glycine = \({mass\ of\ glycine\over molar\ mass\ of\ glycine}\)
\(=\frac{0.1}{0.5}=0.2 \mathrm{~m}\)

Given : C = 0.25 M
T = 370.28 K
R = 0.0821 L atm mol^-1 K^-1
\(\pi\) = ?
\(\pi\) = CRT
= 0.25 x 0.0821 x 370.28
= 7.59 atm.

Add the solute to all the solutions and structures.
a) If the solute dissolves, then that solution is an unsaturated one
b) If the solute settledown at the bottom then that solution is a saturated one
c) If precipitation (crystallisation) occurs then that solution is a super saturated one.

i) Molal depression constant:
Molal freezing point depression constant or Cryoscopic constant is defined as the depression in freezing point for 1 molol solution.
ii) It doesn't depend on the nature of the solute but depends on the nature of the solvent.
Example : K_f for water = 1.86 K kg mol^-1
K_f for benzene = 5.12 K kg mol^-1

Raoult law states that. "in the case of a solution of volatile liquids, the partial vapour pressure of each component (A & B) of the solution is directly proportional to its mole fraction.
According to Raoult's law,
P_A \(\alpha\) X_A 
p_A = K X_A
When xA = 1, k = \({ p }_{ A }^{ 0 }\)
Where \({ p }_{ A }^{ 0 }\) is the vapour pressure of pure component 'A' at the same temperature
"The relative lowering of vapour pressure of an ideal solution containing the nonvolatile solute is equal to the mole fraction of the solute at a given temperature".

Derivation:
when sodium chloride is added to the water, the vapour pressure of the salt solution is lowered. The vapour Pressure of the solution is determined by the number of molecules of the solvent present in the surface at any time and is proportional to the mole fraction of the solvent.
P_solution \(\infty \) X_A
Where X_A is the mole fraction of the solvent
P_solution  = K X_A
When X_A = 1, K = P⁰ solvent
\({ P }_{ solvent }^{ 0 }\) is the partial pressure of pure solvent
P_solution = \({ P }_{ solvent }^{ 0 }\) X_A
\(\frac { { P }_{ solution } }{ { P }_{ solvent }^{ 0 } } ={ X }_{ A }\)
\(1-\frac { { P }_{ solution } }{ { P }_{ solvent }^{ 0 } } 1-{ X }_{ A }\)
\(\frac { { P }_{ solvent }^{ 0 }-{ P }_{ solution } }{ { P }_{ solvent }^{ 0 } } ={ x }_{ B }\)
Where X_B the mole fraction of the solute
( \(\because\) X_A + A_B = 1, X_B = 1 - X_A)
The above expression gives the relative lowering of vapour pressure. Based on this expression, Raoult's law can also be stated as "the relative lowering of vapour pressure of an ideal solution containing, the nonvolatile solute is equal to the mole fraction of the solute at a given temperature".

(i) The pressure of the vapour in equilibrium with its liquid is called vapour pressure of the liquid at the given temperature.
(ii) The ratio of the difference between the vapour pressure of pure solvent and the vapour pressure of a solution to the vapour pressure of pure solvent is called the relative lowering of vapour pressure.
Relative lowering of vapour pressure = \(\frac{\mathbf{P}^{0} \text { solvent }-\mathbf{P}_{\text {soultion }}}{\mathbf{P}_{\text {solvent }}^{0}}\).

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 = \({Theoretical\ molar\ mass\over Experimental\ molar\ mass}={58.5\over 38.75}=1.50\)

W₂ = 2g, W₁ = 75 g
ΔT_f = 0.2 K , K_f = 5.12 K Kg mol^–1
M₂ = ?
\(M_2={K_f\times W_2\times 1000\over \Delta T_f\times W_1}={5.12\times 2\times 1000\over 0.2\times 75}\)
= 682.66 g mol^-1.

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

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

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.

\(P_A^o\) = 10 torr, P_solution = 9 torr
W_A = 20 g  W_B = 1 g
M_A = 200 g mol^-1  M_B = ?
\({\Delta P\over P_A^o}={W_B\times M_A\over M_B\times W_A}\)
\({10-9\over 10}={1\times 200\over M_B\times 20}\)
\(M_B={200\over 20}\times 10=100\ g\ mol^{-1}\)

\(P_{pure\ benzene}^o=\) 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.

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.

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_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.

(a) mass of 1.5 moles of CoCl₂ = 1.5 x 129.9
= 194.85 g
194.85 g anhydrons cobalt chloride is dissolved in water and the solution is make up to one litre in a standard flask.
(b) \(6\%{V\over V}\)  aqeous solution contains 6g of methanol in 100 ml solution.
\(\therefore\) To prepare 500 ml of  \(6\%{V\over V}\) solution of methanol 30g methanol is taken in a 500 ml standard flask and required quantity of water is added to make up the solution to 500 ml.

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.

ppm = \({mass\ of\ dissolved\ solid\over mass\ of\ water}\times 10^6\)
\({5\times 10^{-3}\ g\over 1.05\times 10^{3}\ g}\times 10^6=\) 4.76 ppm.

\(10\%{W\over V}\) 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.

both assertion and reason are true and reason is the correct explanation of assertion

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

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.