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

1. Place the egg in dil. HCl or dil. H₂SO₄ solution.
2.  After sometime , outer shell of egg dissolves.
3. Egg with only semi-permeable membrane is now removed and placed in hypertonic solution.
4. After sometime. the size of the egg gets reduced due to osmosis.
5. Egg is now placed in bottle with narrow neck
6. Add hypotonic solution in the bottle containing egg.
7. Egg regains its shape due to osmosis.

ΔT_b = 100.0832 - 100.0 = 0.0832°C,
K_b = 0.52 K kg mol^-1,
W_A = 250 g, W_B = 3.12 g,
M_B = 137 + 71 = 208 9 mol^-1
Now \(ΔT_b=iK_b\times{W_B\over M_B}\times{1000\over W_A}\)
⇒ \(0.0832 = i \times 0.52 \times{3.12\over4 6.489}\times{1000\over 250}\)
\(⇒ i={0.0832\times208\over 4\times3.12\times0.52}={17.30\over 6.489}=2.66\)
\(\alpha={i-1\over n-1}={2.66-1\over 3-1}={1.66\over 2}=0.83\)
Degree of dissociation (α) = 83%

\(\pi V\)= nRT = \(\frac{{W}^{B}}{{M}^{B}}\times R \times T\)

\(\Rightarrow \)  \(\pi \times1 = \frac{68.4}{342}\times0.0821 \times 300 K \)

\(\Rightarrow \) \(\pi =\frac{24.63}{5} = 4.92 \ atm\)

(i) Phenol is partially soluble in water
(ii) Toluene is insoluble in water
(iii) Formic acid is soluble in water
(iv) Ethylene glycol is soluble in water
(v) Chloroform is insoluble in water
(vi) Pentanol is partially soluble in water.

According to Raoult’s Law,
\(\frac{\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}}=\frac{\mathrm{W}_{\mathrm{B}}}{\mathrm{M}_{\mathrm{B}}} \times \frac{\mathrm{M}_{\mathrm{A}}}{\mathrm{W}_{\mathrm{A}}}\)
\(\mathrm{P}_{\mathrm{A}}^{\circ} \text { (for water) }=1 \cdot 013 \mathrm{bar} ; \mathrm{P}_{\mathrm{S}}=1 \cdot 004 \mathrm{bar} ; \mathrm{W}_{\mathrm{B}}=2 \mathrm{~g} ; \mathrm{W}_{\mathrm{A}}=100-2=98 \mathrm{~g}\)
\(\mathrm{M}_{\mathrm{A}}=18 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\frac{(1 \cdot 013-1 \cdot 004) \text { bar }}{(1 \cdot 004 \text { bar })}=\frac{(2 \mathrm{~g}) \times\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)}{\mathrm{M}_{\mathrm{B}} \times(98 \mathrm{~g})}\)
\(\therefore M_{B}=\frac{(2 g) \times\left(18 g m o l^{-1}\right) \times(1 \cdot 004 \text { bar })}{(0.009 \text { bar }) \times(98 g)}=41 \cdot 0 \mathrm{~g} \mathrm{~mol}^{-1}\)

1 molal solution of solute means 1 mole of solute in 1000g of the solvent.
\(\text { Molar mass of water (solvent) }=18 \mathrm{~g} \mathrm{~mol}^{-1}\)
\(\therefore \text { Moles of water }=\frac{1000}{18}=55.5 \text { moles. }\)
\(\therefore \text { Mole fraction of solute }=\frac{1}{1+55 \cdot 5}=0 \cdot 0177\)
\(\text { Now, } \frac{P^{\circ}-P_{s}}{P^{\circ}}=x_{2}\)
\( \frac{12 \cdot 3-P_{s}}{12 \cdot 3}=0 \cdot 0177 \\ \Rightarrow P_{s}= 12 \cdot 08 \mathrm{kPa} \)

269.27 K

\(\triangle { T }_{ f }={ K }_{ f }\times m\)
\(\\ \Rightarrow \ \triangle { T }_{ f }=5.12 \ K \ \times 0.1\)
\(\\ \Rightarrow \ \triangle { T }_{ f }=0.512 \ K\)
\(\\ i=\frac { Observed\triangle { T }_{ f } }{ Normal\triangle { T }_{ f } } =\frac { 0.256 }{ 0.512 } =\frac { 1 }{ 2 } \)
It shows that acetic acid exists as dimer in benzene.

\(\triangle { T }_{ b }=373.128 \ K-373.0 \ K=0.128 \ K\)
\(\\ \triangle { T }_{ b }={ K }_{ b } \ \times \ m\)
\( \Rightarrow m=\frac { 0.128 }{ 0.52 } =0.246 \ mol/kg\)
\( Now,\triangle { T }_{ f }={ K }_{ f } \ \times \ m=1.86\times0.246=0.457\)
Freezing point = \(273-0.457=272.543 \ K\) 

\(\text { No. of moles of solute, } n_{2}=\frac{100}{140}=\frac{5}{7} \text { mole }\)
\(\text { No. of moles of solvent, } n_{1}=\frac{1000}{180}=\frac{50}{9} \text { mole }\)
Mole fraction of solute
\(x_{2}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{5 / 7}{5 / 7+50 / 9}=0.114\)
\(\text {Mole fraction of solvent, } x_{1}=\left(1-x_{2}\right)=(1-0 \cdot 114)\)
= 0.886
According to Raoult's law
\( P_{A}=x_{A} P_{A}^{0}=0.114 \times P_{A}^{0} \)
\(P_{B}=x_{B} P_{B}^{\circ}=0.886 \times 500=443 \text { torr } \)
\(P_{\text {Total }}=P_{A}+P_{B}\)
\(475=0.114 P_{A}^{\circ}+443\)
\( P_{A}^{\circ}=\frac{475-443}{0 \cdot 114}=280 \cdot 7 \text { torr } \)
\(\therefore P_{A}=0 \cdot 114 \times 280 \cdot 7=32 \text { torr. }\)

Alcohol is polar compound. It can form H-0bound with water molecule. That is why alcohols are miscible with water in all proportion. They also have dipole-dipole interactions due to which there is the force of attraction between alcohol and water molecules.