In C₂H₅OH, i = 1, because it is not an electrolyte.
In Ba₃(PO₄)₂ i = 5; In Na₂SO₄, i = 3;
In KCI, i = 2; In Li₃PO₄, i = 4
Greater the value of i, greater will be ΔT_b' and higher will be b.pt. of solution.C2HsOH, KCI, Na₂SO₄, Li₃PO₄ and Ba₃(PO₄)₂ is the increasing order of b.pt. Greater the value of i, greater will be ΔT_f lower will be freezing point of solution.Therefore, Ba₃(PO₄)₂ Li₃PO₄, Na₂SO₄, KCI and C₂H₅OH is the increasing order of freezing point.

\(m=0.561,\triangle { T }_{ f }=2.9{ 3 }^{ \circ }C,i=?\)
\(\\ K_{ f }=1.8{ 6 }^{ \circ }C \ kg \ { mol }^{ -1 }\)
\(\\ \triangle { T }_{ f }=K_{ f }\times m=1.86\times 0.561=1.04346\)
\( i=\frac { (\triangle { T }_{ f })observed }{ (\triangle { T }_{ f })calculated } =\frac { 2.93 }{ 1.04 } =2.81\)

Molar mass of solute,
 \({ M }_{ B }=\frac { { K }_{ f }\times{ W }_{ B }\times1000 }{ { W }_{ A }\times \triangle { T }_{ f } } \)
\( { W }_{ B }=1.0g\)
\( { W }_{ A }=50.0g\)
\( \triangle { T }_{ f }=0.40K\)
\( { K }_{ f }=5.12 \ K \ kg \ { mol }^{ -1 }\)
\(\\ { M }_{ B }=\frac { 5.12\times1.0\times1000 }{ 50\times0.40 } \)
\(=256 \ g/mol\)

d = 1.9 9 mL^-l, W_B= 99.0 g, M₈ = 98 9 mol^-1,
\(Volume={Mass\over Density}={100\over 1.9}cm3\)
\(Molanty(M) ={W_B\over M_B}\times{1000\over vol.\ of\ solution}\)
\(={99\over 98}\times{1000\over {100\over 1.9}}=19.19M\)

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

(a) Osmosis is the flow of solvent from solution of lower concentration to higher concentration through a semi-permeable membrane.
Osmotic pressure is the excess pressure which must be applied to a solution to prevent the passage of solvent through a semi-permeable memberane.
It has been found experimentally that for n moles of the solute dissolved in V litres of the solution, the osmotic pressure (π) at temperature T is
πV=nRT
Where R is a gas constant.
or \(\pi={n\over V}RT\)
= C RT
Where Cis the molar concentration ofthe solution.For a solution at given tempeature, both R and T are constant, so that
π ∝ C
Thus, osmotic pressure depends upon the molar concentration ofsolution and therefore, is a colligative property.
(b) \(ΔT_b={iK_b\times1000\times W_2\over W_1\times M_2}\)
NaCI dissociates as:
NaCI ⇾ Na⁺ + Cl^-
i = 2
W₂ = 1.5.0g, W₁ = 250.0 g, M₂ = 58.44 g mol^-1 K_b = 0.512 K kg mol^-1
\(\Delta T_b={2\times 0.512\times1000\times15.0\over 250.0\times58.44}\)
= 1.05 °C
Boiling point of solution = 100 + 1.05
= 101.5° C

Molality of sugar solution = \(\frac{W_2\times 1000}{M_2 \times W_1}\)
\(=\frac{5}{342}\times \frac{1000}{95}\)
= 0154kg mol^-1 or 0.154 m
\(\Delta T_f=T_f^{0}-T_f=273.15-271=2.15 K\)
\(\Delta T_f=K_f\times m\)
\(\Rightarrow K_f=\frac{\Delta T_f}{m}=\frac{2.15}{0.154}\)
Molality of glucose solution,
\(\frac{W_2\times 1000}{M_2W_1}=\frac{5}{180}\times \frac{1000}{95}=0292 m\)
\(\therefore \Delta T_{f_{glucose}}=K_f\times m=\frac{2.15}{1.54}\times 292=4.08\)
∴  Freezing point of glucose solution = 273.15 - 4.08 = 269.07 K

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

Acetone-chloroform mixture shows negative deviation from Raoult's law. Hence, for a particular composition, it has lowest vapour pressure and maximum boiling point.

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