The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions. For ideal solutions, ∆H(mixing) = 0 and ∆V (mixing) = 0, e.g. solution of n-hexane and n-heptane, bromoethane and chloroethan etc. In these solutions (binary solutions), A-B type (i.e. solute-solvent) interactions are nearly equal to the A-A (solute-solute) and B-B type (solvent-solvent) interactions.

Henry's law states that the mass of a gas dissolved in given volume of the liquid at constant temperature depends upon the pressure applied.

Henry's law states that the mass of a gas dissolved per unit volume of the solvent is directly proportional to the pressure of the gas in equilibrium with the solution. If m is the mass of the gas dissolved in a unit volume of the solvent and P is the pressure of the gas in equilibrium with the solution, then
\(m\propto p\\ or\ m=K.p\)
(where K is the proportionality constant) The dissolution of a gas in a liquid is exothermic process. Therefore, in accordance with Le Chatelier's principle, with increase in temperature, the equilibrium shifts in the backward direction. As a result, solubility decreases with increase in temperature.
Differences between ideal and non-ideal solutions

Ethylene glycol.

(b) If x_A is mol fraction of A in liquid phase, then x_A (450)+(1 - x_A)(700) = 600 or x_A = 0.4, x_B = 0.6.
Hence, \({ p }_{ A }=0.4\times 450=180\quad mm,{ p }_{ B }=0.6\times 700=420\quad mm,{ y }_{ A }=\frac { 180 }{ 600 } =0.3,{ y }_{ B }=\frac { 420 }{ 600 } =0.7\)

i > 1 when the solute undergoes dissociation in the solution.

In osmosis, net flow of solvent across the semipermeable membrane is from solvent to solution. The flow can be reversed by applying pressure greater than osmotic pressure on the solution. This is called reverse osmosis. It is used for desalination of sea water.

The solubility first first increases upto 32.4^oC (called transition temperature) and then decreases.

(i) Both are polar and hence miscible.
(ii) They form hydrogen bonds with each other. This further increases their miscibility.

Molality = 0.1075, (i) 0.055°C  (ii) 0.046 mm

1.204 x 10²¹ molecules

60 u

0.423

(i) Desalination (i.e., removal of salts from sea water) by reverse osmosis.
(ii) From sea water container to fresh water container.
(iii) Cellulose acetate placed over a suitable suport.
(iv) To remove salts from sea water to obtain drinking water.

The egg placed in pure water will swell, whereas the egg placed in saturated solution of NaCl will shrink.

\(=\frac{W_E}{M_E} \times \frac{1000}{\text { Volume of solution in ml }}=\frac{6}{60} \times \frac{1000}{1000}=0.1 \mathrm{M} \text {. }\)

Molal Elevation Constant (Ebullioscopic Constant:) It is equal to elevation in boiling point of 1 molal solution, i.e. 1 mole of solute is dissolved in 1 kg of solvent. It is also called ebullioscopic constant. The units of k_b is K/m or ^oC/m or K kg mol^-1, where 'm' is molality.

(i) It shows positive deviation.
It is due to weaker interaction between acetone and ethanol than ethanol-ethanol interactions.
(ii) Given : W_B  = 10g, W_s = 100 g, W_A = 90 g, M_B = 180 g.mol and d = 1.2 g/mL
\(M=\frac { Wt%\times density\times 10 }{ Mol.wt } \)
\(M=\frac { 10\times 1.2\times 10 }{ 180 } \)
= 0.66 M or 0.66 mol/L
\(m=\frac { { W }_{ B }\times 1000 }{ { M }_{ B }\times { W }_{ A }(in\quad g) } \)
\(m=\frac { 10\times 1000 }{ 180\times 90 } \)
= 0.61 m or 0.61 mol/kg
(or any other suitable method)

\((i){p_A^0-p_A\over p_A^0}=x_B\Rightarrow1-{p_A\over p_A}={{30\over M_B}\over{90\over 18}}\)
\(\Rightarrow\ \ 1-{2.8\over p_A^0}={6\over M_B}\Rightarrow{2.8\over p_A^0}=1-{6\over M_B}={M_B-6\over M_B}\)
\(\Rightarrow\ {2.8\over p_A^0}={M_B-6\over M_B}\)
Similarly, \(1-{2.9\over p_A^0}={{30\over M_B}\over{108\over 18}}\Rightarrow\ 1-{2.9\over p_A^0}={5\over M_B}\)
\(\Rightarrow\ {2.9\over p_A^0}={M_B-5\over M_B}\)
Dividing (0) by (b), we have
\({2.8\over 2.9}={M_B-6\over M_B-6}\)
=> 2.8 M_B -14 = 2.9 M_B - 17.4
0.1 M_B = 3.4
Molecular weight of
solute = M_B = 34 g mol^-1.
(ii) From (a),
\({2.8\over p_A^0}={34-6\over 34}\Rightarrow\ p_A^0={34\times2.8\over 28}=3.4kpa\)
Vapour pressure of pure water = 3.4 kPa.

The plot of p_(total) dips downwards and therefore, the solution shows negative deviation from ideal behaviour.

(a) (i) Henry's Law: It states that the solubility of a gas in liquid is directly proportional to the pressure of the gas. P = K_Hx, where 'P' is pressure of gas, 'x' is mole fraction of of the gas and K_H is Henry's law constant.
(ii) Boiling Point Elevation Constant (Molal Boiling Point Elevation Constant): It is equal to elevation in boiling point of 1molal solution, i.e. 1mole of solute is dissolved in 1 kg of solvent. The units of K_b is Kim or ⁰C/m or K kg mot".
(b) W_B =? M_B = 36 + 8 + 48 = 92 g mol^-1, W_A = 500 g, ~T_b = 100.42⁰C - 100⁰C = 0.42⁰C
\(\Delta T_b=K_b\times {W_B\over M_B}\times{1000\over W_A}\)
\(\Rightarrow\ 0.42=0.512\times{W_B\over 92}\times{1000\over 500}\)
\(\Rightarrow\ W_B={0.42\times92\over 2\times0.512}={38.64\over 1.024}=37.73g\)

Assume that we have 100 g of solution (one can start with any amount of solution because the results obtained will be the same). Solution will contain 20 g of ethylene glycol and 80 g of water.
Molar mass of C₂H₆O₂ = 12 x 2 + 1 × 6 + 16 × 2 = 62 g mol^-1
\(\text {Moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}=\frac{20 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.322 \mathrm{~mol}\)
\(\text {Moles of water }=\frac{80 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}=4.444 \mathrm{~mol}\)
\(\mathrm{x}_{\text {glycol }}=\frac{\text { moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}}{\text { moles of } \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}+\text { moles of } \mathrm{H}_{2} \mathrm{O}}\)
\(=\frac{0.322 \mathrm{~mol}}{0.322 \mathrm{~mol}+4.444 \mathrm{~mol}}=0.068\)
\(\text {Similarly, } x_{\text {water }}=\frac{4.444 \mathrm{~mol}}{0.322 \mathrm{~mol}+4.444 \mathrm{~mol}}=0.932\)
Mole fraction of water can also be calculated as: 1- 0.068 = 0.932

Molar mass of KCI = 39 + 35.5 = 74.35 g mol^-1
A KCI dissociates completely, number of ions produced are 2.There, Van't Hoff factor, i = 2
Mass of KCI solution = 1000 \(\times\)1.04 = 1040 g
Mass of solvent = 1040 - 74.5 = 965.5 g = 0.9655 kg^1/2
Molality of the solution :
\(\\ \frac { No.\quad of\quad moles\quad of\quad solute }{ Mass\quad of\quad solvent\quad in\quad kg } =\frac { 1\quad mol }{ 0.9655\quad kg } =1.0357\quad m\)
T_b = i \(\times\)K_b \(\times\)m
= 2 \(\times\) 0.52 \(\times\)1.0357 = 1.078 \(°\) C
Therefore, boiling point of solution
= 100 = 1.078 = 101.078\(°\)C

Depression in freezing point is related to the molality, therefore, the molality of the solution with respect to ethylene glycol \(=\frac{\text { moles of ethylene glycol }}{\text { mass of water in kilogram }}\)
\(\text {Moles of ethylene glycol }=\frac{45 \mathrm{~g}}{62 \mathrm{~g} \mathrm{~mol}^{-1}}=0.73 \mathrm{~mol}\)
\(\text {Mass of water in } \mathrm{kg}=\frac{600 \mathrm{~g}}{1000 \mathrm{~g} \mathrm{~kg}^{-1}}=0.6 \mathrm{~kg}\)
\(\text {Hence molality of ethylene glycol }=\frac{0.73 \mathrm{~mol}}{0.60 \mathrm{~kg}}=1.2 \mathrm{~mol} \mathrm{~kg}^{-1}\)
Therefore freezing point depression,
ÄT_f = 1.86 K kg mol^–1 x 1.2 mol kg^–1 = 2.2 K
Freezing point of the aqueous solution = 273.15 K – 2.2 K = 270.95 K

\(\Delta\)T_b = iK_bm
i = 2
\(\Delta\)T_b = i \(\times\)K_b \(\times\)\(\frac { { W }_{ 2 }\times 1000 }{ M\times { W }_{ 1 } } \)
\(=2\times 0.52 \ K \ kg \ { mol }^{ -1 }\times \frac { 4g\times 1000 \ g/kg }{ 120 \ g/mol\times 100g } \)
\(=\frac { 2\times 0.52 }{ 3 } =0.346\ K\)
Boiling point of solution=\(\frac { 373.15\quad K }{ 373\quad K } \)
T_b = \({ \Delta T }_{ b }^{ o }+\Delta { T }_{ b }\)
\(=\frac { 373.15+0.346\quad K }{ 373\quad K+\quad 0.346\quad K } \)
\(=\frac { 373.496\ K }{ 373.346\ K } \)

(i) The cell walls of the vegetables have semi-permeable membrane. When the vegetables have dried, the concentration of water inside becomes low. When the dried or shrinked vegetables are placed in water, water enters into the cells due to osmosis. Hence, they swell and return to the original form.
(ii) The process is osmosis. It is defined as the flow of the solvent molecules from the solvent to the solution or from less concentrated solution to a more concentrated solution through a semi-permeable membrane.
(iii) The process will be accelerated with increase of temperature because osmosis becomes faster with increase of temperature.

0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH₂CONH₂) = 60 g mol^-1
\(\therefore\) 0.25 mole of urea = 0.25 x 60 = 15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
\(\therefore\)2.5 kg of solution contains urea = 15/1.015 x 2.5 = 37 g

According to Raoult’s Law,
\(\frac{\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}}=\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \times \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{A}}}\)
\(\text { Let } \mathrm{P}_{\mathrm{A}}^{\circ}=1 \mathrm{~atm}, \mathrm{P}_{\mathrm{S}}=0 \cdot 8 \mathrm{~atm} ; \mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{S}}=0 \cdot 2 \mathrm{~atm} ; \mathrm{M}_{\mathrm{B}}=40 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{W}_{\mathrm{A}}=114 \mathrm{~g} \)
\(\mathrm{M}_{\mathrm{A}}\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)=114 \mathrm{~g} \mathrm{~mol}^{-1} \)
\(\mathrm{W}_{\mathrm{B}}=\frac{\left(\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{S}}\right)}{\mathrm{P}_{\mathrm{S}}} \times \frac{\mathrm{M}_{\mathrm{B}} \times \mathrm{W}_{\mathrm{A}}}{\mathrm{M}_{\mathrm{A}}}\)
\(=\frac{(0 \cdot 2 \mathrm{~atm})}{(0 \cdot 8 \mathrm{~atm})} \times \frac{\left(40 \mathrm{~g} \mathrm{~mol}^{-1}\right) \times(114 \mathrm{~g})}{\left(114 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=10 \cdot 0 \mathrm{~g}\)

\(\frac { { p }_{ A }^{ \circ }-{ p }_{ A } }{ { p }_{ A }^{ \circ } } ={ X }_{ B }=\frac { \frac { { W }_{ B } }{ { M }_{ B } } }{ \frac { { W }_{ A } }{ { M }_{ A } } +\frac { { W }_{ B } }{ { M }_{ B } } } \)
\(\\ \Rightarrow \frac { 3.165kPa-{ p }_{ A } }{ 3.165 \ kPa } =\frac { \frac { { W }_{ B } }{ { M }_{ B } } }{ \frac { { W }_{ A } }{ { M }_{ A } } } \)
\([\because \frac { { W }_{ B } }{ { M }_{ B } } <<<\frac { { W }_{ A } }{ { M }_{ A } } ]\)
\(\\ Molecular \ weight \ of \ urea=60.05 \ g \ { mol }^{ -1 }\)
\(\\ \Rightarrow 1-\frac { { p }_{ A } }{ 3.165 } =\frac { \frac { 5 }{ 60.05 } }{ \frac { 95 }{ 18 } } \)
\(\\ \Rightarrow 1-\frac { { p }_{ A } }{ 3.165 } =\frac { 5 }{ 60.05 } \times \frac { 18 }{ 95 } =\frac { 18 }{ 1140.95 } \)
\( \Rightarrow \frac { { p }_{ A } }{ 3.165 } =1-\frac { 18 }{ 1140.95 } \)
\(=\frac { 1140.95-18.00 }{ 1140.95 } =\frac { 1122.95 }{ 1140.95 } \)
\(\Rightarrow \ \ { p }_{ A }=\frac { 3.165\times1122.95 }{ 1140.95 }\)
\(=3.115 \ kPa=3.12 \ kPa\)

0.01 M MgCI₂ = 0.03 M particle concentration which is three times in comparison to 0.01 M glucose. Hence, depression will be about three times.

Body temperature of the human body remains constant. Low concentration of oxygen in the blood at altitude is due to low atmospheric pressure.

In case of dissociation, i>1 and in case of association, i>1.

As solute is solid, the solvent, i.e., water will freeze out.

\(\pi =CRT\\ C=0.002\quad mol\quad { dm }^{ -3 }=2\quad mol\quad { m }^{ -3 }\\ R={ 8.314JK }^{ -1 }{ mol }^{ -1 },T=293K\\ \pi =2\times 80314\times 293=4872\quad Pa\)

Higher the solubility, lower is K_H. As CO₂ has maximum solubility, its K_H is lowest.