Solutions

$W_{\text {solute }}=2.5 \times 10^{-3} \mathrm{~kg}$

$ \begin{aligned} &\mathrm{W}_{\text {solvent }}=75 \times 10^{-3} \mathrm{~kg} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =373.535-373.15 \\\\ &=0.385 \mathrm{~K} \\\\ &\mathrm{~K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right) =0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \\\\ &\Delta \mathrm{T}_{\mathrm{b}} =\frac{\mathrm{K}_{\mathrm{b}} \times 10^{3} \times \mathrm{W}_{\text {solute }}}{\mathrm{M}_{\text {solute }} \times \mathrm{W}_{\text {solvent }}} \\\\ &\mathrm{M}_{\text {solute }} =\frac{0.52 \times 10^{3} \times 2.5 \times 10^{-3}}{75 \times 10^{-3} \times 0.385} \\\\ &=45.02 \\\\ & \approx 45 \end{aligned} $

$\Delta T b=k b m$

$ \begin{aligned} &\frac{\left(\Delta T_{b}\right)_{A}}{\left(\Delta T_{b}\right)_{B}}=\frac{\left(k_{b}\right)_{A}}{\left(k_{b}\right)_{B}} \\\\ &=\frac{1}{8}=\frac{x}{y} \\\\ &\therefore y=8 \end{aligned} $

$\pi=\mathrm{i} C R T \quad(\mathrm{i}=1)$

$\pi=\frac{2 \times 1000}{60 \times 10^{3} \times 200} \times .083 \times 300$

$\pi=0.00415 \mathrm{~atm}$

$\pi=415 \mathrm{~Pa}$

$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{b}} \mathrm{m}$

$ \begin{aligned} &\mathrm{i}=\frac{0.24 \times 99.5 \times 74.6}{1.80 \times 0.5 \times 1000} \\\\ &=1.98 \\\\ &\alpha=\frac{\mathrm{i}-1}{\mathrm{n}-1}=\frac{0.98}{1}=0.98 = 98 \,\% \end{aligned} $

According to Henry's law, partial pressure of a gas is given by

$P_{g}=\left(K_{H}\right) X_{g}$

where $X_{g}$ is mole fraction of gas in solution

$0.835=1.67 \times 10^{3}\left(\mathrm{X}_{\mathrm{CO}_{2}}\right)$

$\mathrm{X}_{\mathrm{CO}_{2}}=5 \times 10^{-4}$

Mass of $\mathrm{CO}_{2}$ in $1 \mathrm{~L}$ water $=1221 \times 10^{-3} \mathrm{~g}$

$7.47=\mathrm{C} \times 0.083 \times 300$

$(\pi=\mathrm{CRT})$

(Where C represents the concentration of glucose solution and $\pi$ represents osmotic pressure)

$\mathrm{C}=\frac{7.47}{0.083 \times 300}\left(\mathrm{~mol} \mathrm{~L}^{-1}\right)$

which in $\mathrm{gm} / \mathrm{L}=\frac{7.47}{0.083 \times 300} \times 180$

$ =54 \mathrm{gm} / \mathrm{l} $

With benzene as solvent

$\Delta$T_b = i K_b m

$\Delta$T_b = ${1 \over 2}$ $\times$ 2.6 $\times$ ${{1.22/{M_w}} \over {100/1000}}$ .... (1)

With Acetone as solvent

$\Delta$T_b = i K_b m

0.17 = 1 $\times$ 1.7 $\times$ ${{1.22/{M_w}} \over {100/1000}}$ ..... (2)

(1) / (2)

${{\Delta {T_b}} \over {0.17}} = {{{1 \over 2} \times 2.6 + {{1.22/{M_w}} \over {100/1000}}} \over {1 \times 1.7 \times {{1.22/{M_w}} \over {100/1000}}}}$

$\Delta$T_b = ${{0.26} \over 2}$

$\Delta$T_b = 13 $\times$ 10^$-$2

$\Rightarrow$ x = 13

Molality = ${{\left( {{{40} \over {180}}} \right)mol} \over {0.2\,Kg}} = \left( {{{10} \over 9}} \right)$ molal

$ \Rightarrow \Delta {T_f} = {T_f} - {T_f}' = 1.86 \times {{10} \over 9}$

$ \Rightarrow {T_f}' = 273.15 - 1.86 \times {{10} \over 9}$

= 271.08 K

$ \simeq $ 271 K (nearest - integer)

Let mass of water initially present = x gm

$\Rightarrow$ Mass of sucrose = (1000 $-$ x) gm

$\Rightarrow$ moles of sucrose = $\left( {{{1000 - x} \over {342}}} \right)$

$ \Rightarrow 0.75 = {{\left( {{{1000 - x} \over {342}}} \right)} \over {\left( {{x \over {1000}}} \right)}} \Rightarrow {x \over {1000}} = {{1000 - x} \over {342 \times 0.75}}$

$\Rightarrow$ 256.5x = 10⁶ $-$ 1000x

$\Rightarrow$ x = 795.86 gm

$\Rightarrow$ moles of sucrose = 0.5969

New mass of H₂O = a kg

$ \Rightarrow 4 = {{0.5969} \over a} \times 1.86 \Rightarrow $ a = 0.2775 kg

$\Rightarrow$ ice separated = (795.86 $-$ 277.5) = 518.3 gm

k_f = 1.86 kg/mol

T$_f^o$ = 273 K

solvent : H₂O(625 g)

Solute : 83 g of ethylene glycol

$\Rightarrow$ $\Delta$T_f = k_f $\times$ m

$\Rightarrow$ $\left( {T_f^o - T_f^1} \right) = 1.86 \times {{83/62} \over {625/1000}}$

$ \Rightarrow 273 - T_f^1 = {{1.86 \times 83 \times 1000} \over {62 \times 625}} = {{154380} \over {38750}}$

$ \Rightarrow 273 - T_f^1 = 4$

$ \Rightarrow T_f^1 = 259$ K

As 0.1 M C₂H₅OH is non-dissociative and rest all salt given are electrolyte so in each case effective molarity > 0.1 so each will have lower freezing point.

Now, $i = {{final\,moles} \over {initial\,moles}} = {{0.25a + 0.5a + 0.5a} \over {0.5a + 0.5a}}$

= 1.25 = 125 $\times$ 10^$-$2

$\pi$ = CRT;

$\pi$ = osmotic pressure

C = molarity

T = Temperature of solution

let the molar mass be M gm/mol

2.42 $\times$ 10^$-$3 bar = ${{\left( {{{1.46g} \over {Mgm/mol}}} \right)} \over {0.1l}} \times \left( {{{0.083l - bar} \over {mol - K}}} \right) \times (300K)$

$\Rightarrow$ M = 15.02 $\times$ 10⁴ g/mol

$\Delta$T_b = K_b $\times$ molality

0.60 = 5 $\times$ $\left( {{{3/M} \over {100/100}}} \right)$

M = 250

From Henry's law

P_gas = K_H.X_gas

0.835 = 1.67 $\times$ 10³ $\times$ ${{n(C{O_2})} \over {{{0.9 \times 1000} \over {18}}}}$

$ \Rightarrow $ n(CO₂) = 0.025

Millimoles of CO₂ = 0.025 $\times$ 1000 = 25

x_A = 0.6

P_T = x_AP_A^o + x_BP_B^o

= 0.6 $\times$ 90 + 0.4 $\times$ 15

= 54 + 6 = 60

x_AP_A^o = y_AP_T

0.6 $\times$ 90 = y_A(60)

$\Rightarrow$ y_A = 0.9

y_B = 0.1 = 1 $\times$ 10^$-$1

$\therefore$ x = 1

Vapour pressure of pure benzene, $p_A^o$ = 70 Torr

Vapour pressure of pure methyl benzene, $p_B^o$ = 20 Torr

This mixture is equimolar, so number of moles of benzene,

n_A = number of moles of methyl benzene, n_B

Mole fraction of benzene in vapour phase, y_A = ${{{p_A}} \over {{p_T}}}$ .... (i)

where p_A is pressure of benzene in mixture.

${p_A} = \mathop {p_A^o{\chi _A}}\limits_{Mole\,fraction} = p_A^o \times {{{n_A}} \over {{n_A} + {n_B}}} = p_A^o \times {{{n \over A}} \over {2{n \over A}}} = {{p_A^o} \over 2}$

p_T is total pressure,

${p_T} = p_A^o{\chi _A} + p_B^o{\chi _B} = {p_A} + {p_B}$ (pressure of methyl benzene)

${p_T} = {{p_A^o} \over 2} + {{p_B^o} \over 2} = {{p_A^o + p_B^o} \over 2}$

Putting in above equation (i),

${y_A} = {{{{70} \over 2}} \over {{{(70 + 20)} \over 2}}} = {{70} \over {90}}$

y_A = 0.78 = 78 $\times$ 10^$-$2

$\Delta$T_b = Boiling point of the solution $-$ Boiling point of the pure solvent

= 100.52 $-$ 100

= 0.52

$ \therefore $ $\Delta$T_b = K_b (iM)

$ \Rightarrow $ 0.52 = i $\times$ 0.52 $\times$ 2

$ \Rightarrow $ i = ${1 \over 2}$

We know,

i = 1 + $\left( {{1 \over n} - 1} \right)\beta $

here, $\beta$ = degree of dimerization

n = number of particle associated

n = 2 for dimerization

$ \therefore $ ${1 \over 2} = 1 + \left( {{1 \over 2} - 1} \right)\beta $

$ \Rightarrow \beta = 1$

$ \therefore $ % association = 100

$\Delta$T_f = K_f (im)

$ \Rightarrow $ 3.885 = i $\times$ 1.85 $\times$ 2

$ \Rightarrow $ i = 1.05

Also, we know,

i = 1 + (n $-$ 1) $\alpha$

here n = number of particle obtained upon the dissociation of one particle.

$HA\rightleftharpoons H^{+}+A^{-} $

here from one particle HA we get two particle H⁺ and A^$-$.

$ \therefore $ n = 2

So, i = 1 + (2 $-$ 1)$\alpha$

$ \Rightarrow $ 1.05 = 1 + $\alpha$

$ \Rightarrow $ $\alpha$ = 0.05 = 50 $\times$ 10^$-$3

Effective molality = 0.6 + 1.6 + 0.4 = 2.6 m

As elevation in boiling point is a colligative property which depends on the amount of solute. So, to have same boiling point, the molality of two solutions should be same.

Molality of non-electrolyte solution = molality of ${K_4}[Fe{(CN)_6}]$ = 2.6 m

Now, 18.1 weight per cent solution means 18.1 g solute is present in 100 g solution and hence, (100 $-$ 18.1) = 81.9 g water.

$Molality = {{(Mass\,of\,solute/Molar\,mass\,of\,solute)} \over {Mass\,of\,solvent\,in\,kg}} \times 1000$

Now, $2.6 = {{18.1/M} \over {81.9/1000}}$

where, M is the molar mass of non-electrolyte solute

Molar mass of solute, M = 85

The oxygen dissolved in water has a partial pressure of 20 kPa in the vapor phase above the water. To find the molar solubility of oxygen in water, we use Henry's Law, which states:

$ \text{Partial pressure} (P_g) \propto \text{Solubility} $

This can be mathematically expressed as:

$ P_g = K_H \times \text{Solubility} $

Where:

$ P_g $ is the partial pressure of oxygen, given as 20 kPa.

$ K_H $ is Henry's law constant for $ \text{O}_2 $, provided as $ 8.0 \times 10^4 \, \text{kPa} $.

Substituting the given values into the equation:

$ 20 \times 10^3 = (8.0 \times 10^4) \times \text{Solubility} $

Solving for solubility:

$ \text{Solubility} = \frac{20 \times 10^3}{8.0 \times 10^4} $

$ \text{Solubility} = 0.25 \times 10^{-1} $

$ \text{Solubility} = 25 \times 10^{-5} \, \text{mol dm}^{-3} $

Thus, the molar solubility of oxygen in water rounds to 25 × 10^-5 mol dm^-3.

${X_A} = {1 \over {1 + 2}} = {1 \over 3}$

${X_B} = {2 \over 3}$

$P_A^o = 21kPa$

$P_B^o = 18kPa$

${P_{total}} = P_A^o{X_A} + P_B^o{X_B}$

$ = 21 \times {1 \over 3} + 18 \times {2 \over 3}$

$ = 7 + 12$

$ = 19 kPa$

AB₂ $ \leftrightharpoons $ A⁺ + 2B^$-$

$ \therefore $ For AB₂, n = 3

i = 1 + (n $-$ 1)$\alpha$

= 1 + (3 $-$ 1) $\times$ 0.1

= 1.2

Now, $\Delta$T_b = K_b (im)

$ \Rightarrow $ T_b $-$ T$_b^o$ = 1.2 $\times$ 0.5 $\times$ 10

$ \Rightarrow $ T_b $-$ 100 = 6

$ \Rightarrow $ T_b = 106

$\underset{\text{Benzoic acid}}{n \mathrm{PhCOOH}} \stackrel{\text { Association }}{\longrightarrow}(\mathrm{PhCOOH})_n$

Assuming $100 \%$ association ( $\alpha=1$ ),

$ \Rightarrow i=1-\alpha\left(1-\frac{1}{n}\right)=\frac{1}{n}[\because \alpha+1] $

Now, $\Delta T_f=K_f \times m \times i$

$ 0-(0.93)=1.86 \times \frac{w_B \times 1000}{w_A \times M_B} \times \frac{1}{n} $

$\left[\because w_B=\right.$ mass of $\mathrm{PhCOOH}=12.2 \mathrm{~g}$

$w_A=$ mass of $\mathrm{H}_2 \mathrm{O}=100 \mathrm{~g}$

$M_B=$ molar mass of $\left.\mathrm{PhCOOH}\right]$

$=122 \mathrm{~g} \mathrm{~mol}^{-1}$

$=1.86 \times \frac{12.2 \times 1000}{100 \times 122} \times \frac{1}{n}$

$ \begin{aligned} \Rightarrow n &=\frac{1.86 \times 12.2 \times 1000}{0.93 \times 100 \times 122}=2 \end{aligned} $

$\therefore$ Number of benzoic acid molecules associated, $n=2$

moles of SO₂ = ${{224} \over {22400}}$ = 0.01

moles of NaOH = molarity × volume (in litre)
= 0.1 × 0.1
= 0.01 moles

The balanced equation is

SO₂ + 2NaOH $ \to $ Na₂SO₃ + H₂O

$ \therefore $ Here NaOH is limiting Reagent.

2 mole NaOH produces 1 mole Na₂SO₃

0.01 mole NaOH produces ${1 \over 2}$ $ \times $ 0.01 mole Na₂SO₃

$ \therefore $ Moles of Na₂SO₃ = 0.005 mole

Na₂SO₃ $ \to $ 2Na⁺ + SO₃^2-

van’t Hoff factor (i) = 3

Moles of H₂O = ${{36} \over {18}}$ = 2 moles

Accoding to Relative Lowering of Vapour :

${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}} + i{n_{N{a_2}C{O_3}}}}}$

[ ${{n_{N{a_2}C{O_3}}}}$ << ${{n_{{H_2}O}}}$ ]

${{P_{{H_2}O}^o - {P_S}} \over {P_{{H_2}O}^o}} = {{i{n_{N{a_2}C{O_3}}}} \over {{n_{{H_2}O}}}}$

$ \Rightarrow $ ${{24 - {P_S}} \over {24}} = {{3 \times 0.005} \over 2}$

$ \Rightarrow $ 24 – P_S = 0.18

$ \Rightarrow $ P_S = 23.82

Lowering in pressure ($\Delta $P) = 0.18 mm of Hg

= 18 × 10^–2 mm of Hg

As AB is a binary electrolyte,

$\therefore$ AB $\rightleftharpoons$ A⁺ + B^$-$, n = 2

$i = 1 + \alpha (n - 1) = 1 + {{75} \over {100}}(2 - 1) = 1.75$

Given, $\Delta$T_b = 2.5 K

K_b = 0.52 K kg mol^$-$1

$\therefore$ $\Delta$T_b = K_b $\times$ m $\times$ i

$ \Rightarrow m = {{\Delta {T_b}} \over {{K_b} \times i}} = {{2.5} \over {0.52 \times 1.75}}$

$ = 2.74 \simeq 3$ mol/kg

${A_2}{B_3}\buildrel {} \over \longrightarrow 2{A^{ + 3}} + 3{B^{ - 2}}$

No. of ions = 2 + 3 = 5

i = 1 + (n $-$ 1) $\alpha $

= 1 + (5 $-$ 1) $\times$ 0.6

= 1 + 4 $\times$ 0.6 = 1 + 2.4 = 3.4

$\Delta {T_b} = {K_b} \times m \times i = 0.52 \times 1 \times 3.4 = 1.768^\circ $ C

$\Delta {T_b} = {({T_b})_{solution}} - {[{({T_b})_{{H_2}O}}]_{Solution}}$

$1.768 = {({T_b})_{solution}} - 100$

${({T_b})_{solution}} = 101.768^\circ $ C

$ = 375$ K

Pure solvent : C₆H₆(l)

Given,

$T_f^o = 5.5^\circ C$

${K_f} = 5.12^\circ C/m \Rightarrow m = 200g$

${m_{solute}} = 10g$

Molar mass of solute ${C_4}{H_{10}} = 12 \times 4 + 10 = 58$

Solute (C₄H₁₀) is non-dissociative;

$\therefore$ i = 1

$\therefore$ $\Delta {T_f} = i{K_f}\,m$

$ \Rightarrow (T_f^o - T_f^1) = 1 \times 5.12 \times {{(10/58)} \over {(200/1000)}}$

$5.5 - T_f^1 = {{5.12 \times 5 \times 10} \over {58}} \Rightarrow T_f^1 = 1.086^\circ C$

or $T_f^1 \approx 1^\circ C$

Here, $i = {{Final\,moles} \over {Initial\,moles}}$

i = 1 $-$ $\alpha$ + $\alpha$ + $\alpha$ $\Rightarrow$ i = 1 + $\alpha$

Formula used for freezing point; $\Delta$T_f = i K_fm

Here, K_f = freezing constant of H₂O

K_f(H₂O) = 1.86 K kg mol^$-$1

m = molarity

$\Delta$T_f = i K_fm

0.5 = (1 + $\alpha$) (1.86)${{(9.45/94.5)} \over {(500/1000)}}$

${5 \over {3.72}}$ = 1 + $\alpha$ $\Rightarrow$ ${5 \over {3.72}}$ $-$ 1 = $\alpha$

$\Rightarrow$ $\alpha$ $ = {{1.28} \over {3.72}} = {{32} \over {93}} = 0.344$

Percentage of dissociation = 34.4%

Molality of CaCl₂ solution = 0.05 m

$\Delta $T_b = i K_b m = 3 × K_b × 0.05 = 0.15 K_b

Molality of CrCl₃.xNH₃ = 0.10 m

$\Delta $T_b^' = i K_b $ \times $ 0.10

Given, $\Delta $T_b^' = 2$\Delta $T_b

$ \Rightarrow $ i K_b $ \times $ 0.10 = 0.15 K_b

$ \Rightarrow $ i = 3

Co-ordination number of Cr is 6.

[Cr(NH₃ )_x.Cl_6-x]Cl_3-6+x $ \to $ [Cr(NH₃ )_xCl] + (3-6+x)Cl^–

i = 3 = 1 + 3 - 6 + x

$ \Rightarrow $ x = 5

For NaCl : $\pi $₁ = 0.1 atm, v₁ = 1 L

For Glucose: $\pi $₂ = 0.2 atm, v₂ = 2 L

$\pi $_mix = ${{{\pi _1}{v_1} + {\pi _2}{v_2}} \over {{v_1} + {v_2}}}$

= ${{0.1 \times 1 + 0.2 \times 2} \over {1 + 2}}$

= ${{0.5} \over 3}$ = ${{500} \over 3} \times {10^{ - 3}}$

= 167 $ \times $ 10^-3 atm

Let X₁ and P$P_1^o$ are the mole fraction and vapour pressure of n-hexane in solution and X₂ and $P_2^o$ are the mole fraction and vapour pressure of n-heptane in solution then

550 = $P_1^o$$ \times $${1 \over 4}$ + $P_2^o$$ \times $${3 \over 4}$

$ \Rightarrow $ 2200 = $P_1^o$ + 3$P_2^o$ ....(1)

On addition of 1 more mole of n-heptane

560 = $P_1^o$$ \times $${1 \over 5}$ + $P_2^o$$ \times $${4 \over 5}$

$ \Rightarrow $ 2800 = $P_1^o$ + 4$P_2^o$ ....(2)

From (1) and (2),

$P_1^o$ = 400, $P_2^o$ = 600

Let molar mass of protein A = x g/mol

Let molar mass of protein B = y g/mol

$\pi $_A = osmotic pressure of protein A = ${{0.73} \over x} \times {{1000} \over {250}} \times RT$

$\pi $_B = osmotic pressure of protein B = ${{1.65} \over y} \times {1 \over 1} \times RT$

Given $\pi $_A = $\pi $_B

$ \Rightarrow $ ${{0.73} \over x} \times {{1000} \over {250}} \times RT$ = ${{1.65} \over y} \times {1 \over 1} \times RT$

$ \Rightarrow $ ${x \over y}$ = ${{0.73} \over {0.25 \times 1.65}}$ = 1.77

= 177 $ \times $ 10^-2

For aqueous solution

$\Delta $T_f = K_f.m = 2 × 0.5

$ \therefore $ Temperature of solution = –1°C = 272 K

$ \therefore $ Final volume of ideal gas = ${{nRT} \over P}$

= ${{0.1 \times 0.08 \times 272} \over 1}$

= 2.18 L

$\Delta $T_f = 0.2^o C

$ \therefore $ $\Delta $T_f = ik_f.m

i = 2 for NaCl

$ \therefore $ 0.2 = 2$ \times $2$ \times $${{{W_{NaCl}} \times 1000} \over {58.5 \times 600}}$

$ \Rightarrow $ W_NaCl = ${{58.5 \times 600 \times 0.2} \over {4 \times 1000}}$ = 1.76 g

Density of the solution:

$ \text{Density} = 1.00 \, \text{g cm}^{-3} = 1000 \, \text{kg m}^{-3} $

Height of the solution column ($h$):

$ h = 2 \, \text{cm} = 2 \times 10^{-2} \, \text{m} $

Acceleration due to gravity ($g$):

$ g = 10 \, \text{m s}^{-2} $

Calculating osmotic pressure ($\pi$):

Osmotic pressure is given by the equation:

$ \pi = h \cdot \rho \cdot g $

Substituting the known values:

$ \pi = 2 \times 10^{-2} \times 1000 \times 10 = 200 \, \text{N m}^{-2} $

Relating osmotic pressure to concentration and molar mass:

The equation for osmotic pressure in terms of concentration and molar mass is:

$ \pi = cRT $

where $c = \frac{2000}{M}$ as the concentration in g/dm$^3$ needs to be converted to mol/dm$^3$ by dividing by the molar mass $M$. Incorporating the given temperature and universal gas constant, we have:

$ 200 = \left(\frac{2000}{M}\right) \cdot 8.3 \cdot 300 $

Solving for the molar mass $M$:

Rearrange and solve the equation to find $M$:

$ M = 24900 \, \text{g mol}^{-1} = 2.49 \times 10^4 \, \text{g mol}^{-1} $

Thus, the value of $X$ is $2.49$.

To determine the elevation of the boiling point for the solutions in both vessels, we need to use the concept of boiling point elevation. The boiling point elevation is given by the formula:

$ \Delta T_b = i \cdot K_b \cdot m $

Where:

• $\Delta T_b$ is the boiling point elevation
• $i$ is the van't Hoff factor
• $K_b$ is the ebullioscopic constant (which is the same for both solutions since they are dissolved in water)
• $m$ is the molality of the solution

We need to compare the boiling point elevations for both solutions in Vessel-1 and Vessel-2.

For Vessel-1, let the boiling point elevation be $\Delta T_{b1}$, and for Vessel-2, let it be $\Delta T_{b2}$.

Let's denote the molar mass of solute $Y$ as $M_Y$ g/mol. According to the problem, the molar mass of solute $X$ is 80% of that of $Y$, i.e.,

$ M_X = 0.8 \cdot M_Y $

The van't Hoff factor for $X$ is 1.2 times that of $Y$, i.e.,

$ i_X = 1.2 \cdot i_Y $

The molality $m$ of each solution can be calculated using the formula:

$ m = \dfrac{w_2}{M \cdot w_1} $

Thus, the molality of $X$ and $Y$ solutions are:

For solute $X$:

$ m_X = \dfrac{w_2}{M_X \cdot w_1} = \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} $

For solute $Y$:

$ m_Y = \dfrac{w_2}{M_Y \cdot w_1} $

Now, substituting the values into the boiling point elevation formula, we get for Vessel-1:

$ \Delta T_{b1} = i_X \cdot K_b \cdot m_X = (1.2 \cdot i_Y) \cdot K_b \cdot \left( \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} \right) $

For Vessel-2:

$ \Delta T_{b2} = i_Y \cdot K_b \cdot m_Y = i_Y \cdot K_b \cdot \left( \dfrac{w_2}{M_Y \cdot w_1} \right) $

To find the ratio of $\Delta T_{b1}$ to $\Delta T_{b2}$:

$ \dfrac{\Delta T_{b1}}{\Delta T_{b2}} = \dfrac{(1.2 \cdot i_Y) \cdot K_b \cdot \left( \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} \right)}{i_Y \cdot K_b \cdot \left( \dfrac{w_2}{M_Y \cdot w_1} \right)} $

Simplifying the ratio:

$ \dfrac{\Delta T_{b1}}{\Delta T_{b2}} = \dfrac{1.2}{0.8} = 1.5 $

This means the elevation of the boiling point for the solution in Vessel-1 is 150% of that in Vessel-2. Thus, the elevation of boiling point for solution in Vessel-1 is 150% of the elevation of the boiling point for the solution in Vessel-2.

1. Find the weight of urea in the 0.2 molal solution :
   
   Given a 0.2 molal solution, there are 0.2 moles of urea in 1000 g of solvent. The weight of urea is thus $0.2 \, \text{mol} \times 60 \, \text{g/mol} = 12 \, \text{g}$.




2. Find the weight of the solution :
   
   The total weight of the solution is the weight of the solvent plus the weight of the solute, or $1000 \, \text{g} + 12 \, \text{g} = 1012 \, \text{g}$.




3. Find the volume of the solution :
   
   Given the density of the solution, we can find its volume by dividing the total weight of the solution by the density: $\frac{1012 \, \text{g}}{1.012 \, \text{g/mL}} = 1000 \, \text{mL}$.




4. Find the amount of urea in 50 mL of the 0.2 molal solution :
   
   If 1000 mL of the solution contains 0.2 moles of urea, then 50 mL of the solution contains $\frac{0.2 \, \text{mol} \times 50 \, \text{mL}}{1000 \, \text{mL}} = 0.01 \, \text{mol}$ of urea.




5. Find the amount of urea in the 250 mL solution :
   
   The 250 mL solution contains $0.06 \, \text{g}$ of urea, or $\frac{0.06 \, \text{g}}{60 \, \text{g/mol}} = 0.001 \, \text{mol}$.




6. Find the total concentration of the solution :
   
   After mixing, the total volume of the solution is $50 \, \text{mL} + 250 \, \text{mL} = 300 \, \text{mL}$, and the total amount of urea is $0.01 \, \text{mol} + 0.001 \, \text{mol} = 0.011 \, \text{mol}$.
   
   So, the concentration of the solution is $\frac{0.011 \, \text{mol}}{300 \, \text{mL}} \times 1000 \, \text{mL/L} = 0.0366 \, \text{M}$.




7. Find the osmotic pressure of the solution :
   
   Finally, the osmotic pressure $\pi$ of the solution can be found using the formula $\pi = CRT$, where $C$ is the concentration, $R$ is the gas constant, and $T$ is the temperature. Substituting the given and calculated values,
   
   we have $\pi = 0.0366 \, \text{M} \times 62 \, \text{L Torr K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} = 682 \, \text{Torr}$.

Vapour pressure of solution $\left(\mathrm{P}_{\mathrm{A}}\right)$ $ =59.724 \mathrm{~mm} \text { of } \mathrm{Hg} $

Vapour pressure of pure water $\left(\mathrm{P}_{\mathrm{A}}^{\circ}\right)$ $ =60.000 \mathrm{~mm} \text { of } \mathrm{Hg} $

Also, $0.1 \mathrm{~mol}$ of an ionic solid is dissolved in $1.8 \mathrm{~kg}$ of water and salt remains $90 \%$ dissocated in the solution.



So, total number of moles $=0.01+0.09 a$ of non-volatile particles.

Now, mass of water $=1.8 \mathrm{~kg}=1.8 \times 1.8 \times 1000 \mathrm{~g}$

Molar mass of water $=18 \mathrm{~g}$

Moles of water $=\frac{1.8 \times 1000}{18}=100$ moles

Using the colligative property, relative lowering in vapour pressure,

$ \frac{\mathrm{P}_{\mathrm{A}}^{\circ}-\mathrm{P}_{\mathrm{A}}}{\mathrm{P}_{\mathrm{A}}^{\circ}}=x_{\mathrm{A}} $

$ \Rightarrow $ $\frac{60-59.724}{60} =\frac{0.01+0.09 a}{100} $

$ \Rightarrow $ $ \frac{0.276}{60} =\frac{0.01+0.09 a}{100} $

$ \Rightarrow $ $ \frac{27.6}{60} =0.01+0.09 a $

$ \Rightarrow $ $ 0.46 =0.01+0.09 a $

$ \Rightarrow $ $ 0.09 a =0.45 $

$ \Rightarrow $ $ a =\frac{0.45}{0.89} $

$ \Rightarrow $ $ a =5 $

So, the number of ions present per formula unit of the ionic salt is 5 .

$ \begin{aligned} & \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \underset{0.1 ~ m}{\mathrm{Ag}^{+}(\mathrm{aq})}+\underset{0.1 ~m}{\mathrm{NO}_3^{-}(\mathrm{aq})}\\\\ & \Delta \mathrm{T}_{\mathrm{b}}=0.2 \times 0.5 \\\\ & =0.1^{\circ} \mathrm{C}=0.1 \mathrm{~K} \end{aligned} $

Boiling point of solution $=100.1^{\circ} \mathrm{C} =\mathrm{X} $

$ \begin{aligned} & \mathrm{AgNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{NO}_3^{-}(\mathrm{aq}) \\\\ & 0.05 \mathrm{~m} \quad\quad\quad\quad\quad 0.05 \mathrm{~m} \quad\quad 0.05 \mathrm{~m} \\\\ & \mathrm{BaCl}_2(\mathrm{aq}) \longrightarrow \mathrm{Ba}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) \\\\ & 0.05 \mathrm{~m} \quad \quad\quad\quad\quad0.05 \mathrm{~m} \quad\quad 0.1 \mathrm{~m} \end{aligned} $

$\mathrm{Ag}^{+}$and $\mathrm{Cl}^{-}$combine to form $\mathrm{AgCl}$ precipitate

$ \begin{array}{cccc} & \mathrm{Ag}^{+}(\mathrm{aq})+ & \mathrm{Cl}^{-}(\mathrm{aq}) \longrightarrow & \mathrm{AgCl}(\mathrm{s}) \\\\ \mathrm{t}=0 & 0.05 \mathrm{~m} & 0.1 \mathrm{~m} \\\\ \mathrm{t}=\infty & 0 & 0.05 \mathrm{~m} \end{array} $

In final solution total concentration of all ions :

$ \begin{aligned} & {\left[\mathrm{Cl}^{-}\right]+\left[\mathrm{NO}_3^{-}\right]+\left[\mathrm{Ba}^{2+}\right]=0.05+0.05+0.05} \\\\ & \begin{aligned} \Delta \mathrm{T}_{\mathrm{b}} & =0.5 \times 0.15 \\\\ & =0.15 \mathrm{~m} \\\\ & =0.075^{\circ} \mathrm{C} \end{aligned} \end{aligned} $

B.P. of solution ' $\mathrm{B}$ ' $=100.075^{\circ} \mathrm{C}$

B.P. of solution ' $\mathrm{A}^{\prime}=100.1^{\circ} \mathrm{C}$

$|y|=100.1-100.075$

$ =0.025=2.5 \times 10^{-2} $

Using Raoult's law equation for a mixture of volatile liquids.

${P_T} = p_A^o{\chi _A} + p_B^o{\chi _B}$

$0.3 = 0.25\chi p_A^o + 0.75\chi p_B^o$ ....(i)

$0.4 = 0.5\chi p_A^o + 0.5\chi p_B^o$ ....(ii)

By solving equation (i) and (ii)

$p_A^o$ = 0.6 bar and $p_B^o$ = 0.2 bar

Thus, the vapour pressure of pure liquid B in bar is 0.2.

When 0.5 g of non-volatile solute dissolve into 39 gm of benzene then relative lowering of vapour pressure occurs. Hence, vapour pressure decreases from 650 mmHg to 640 mmHg.

Given, vapour pressure of solvent (p$^\circ $) = 650 mmHg

Vapour pressure of solution (p_s) = 640 mmHg

Weight of non-volatile solute = 0.5 g

Weight of solvent (benzene) = 39 g

From relative lowering of vapour pressure,

${{p^\circ - {p_s}} \over {p^\circ }} = {x_{Solute}} = {{{n_{solute}}} \over {{n_{solute}} + {n_{solvent}}}}$

${{650 - 640} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + {{39} \over {78}}}}$

${{10} \over {650}} = {{{{0.5} \over {molar\,mass}}} \over {{{0.5} \over {molar\,mass}} + 0.5}}$

$0.5 + 0.5 \times molar\,mass = 65 \times 0.5$

$ \therefore $ Molar mass of solute = 64 g

From molal depression of freezing point,

$\Delta {T_f} = {K_f} \times molality$

$ = {{{K_f} \times {w_{solute}}} \over {{{(MW)}_{solute}} \times {w_{solvent}}}}$

$ \Rightarrow $ $ \Delta {T_f} = 5.12 \times {{0.5 \times 1000} \over {64 \times 39}} $

$\Rightarrow \Delta {T_f} = 1.02K$

We know,

${p_{Total}} = p_A^o \times {\chi _A} + p_B^o \times {\chi _B}$

and for equimolar solutions ${\chi _A} = {\chi _B} = 1/2$

Given, ${p_{total}} = 45$ torr for equimolar solution and $p_A^o = 20$ torr

So, $45 = p_A^o \times {1 \over 2} + p_B^o \times {1 \over 2} = {1 \over 2}(p_A^o + p_B^o)$

or $p_A^o + p_B^o = 90$ torr ....... (i)

But we know $p_A^o = 20$ torr

So, $p_B^o = 90 - 20 = 70$ torr (From Eq. (i))

Now, for the new solution from the same formula

${p_{total}} = 22.5$ torr

$22.5 = 20{\chi _A} + 70(1 - {\chi _A})$ (As ${\chi _A} + {\chi _B} = 1$)

or $22.5 = 70 - 50{\chi _A}$

So, ${\chi _A} = {{70 - 22.5} \over {50}} = 0.95$

Thus, ${\chi _B} = 1 - 0.95 = 0.05$ (as $\chi$_A + $\chi$_B = 1)

Hence, the ratio

${{{\chi _A}} \over {{\chi _B}}} = {{0.95} \over {0.05}} = 19$

For solvent X, $\Delta$T_bX = K_bX m $\Rightarrow$ 2 = 2K_bX m ..... (1)

For solvent Y, $\Delta$T_bY = K_bY m $\Rightarrow$ 1 = 2K_bY m ..... (2)

Dividing Eq. (1) by Eq. (2), we get

${{{K_{bX}}} \over {{K_{bY}}}} = 2$

After adding solute S, molality is the same for both solutions.

Dimerisation is a association property. And for dimerisation van't Hoff factor

i = 1 + $\left( {{1 \over n} - 1} \right)\alpha $

And for dimerisation n = 2

$ \therefore $ For solvent X, van't Hoff factor $i = 1 - {\alpha \over 2}$

$\Delta {T_{bX}} = i{K_{bX}}m = \left( {1 - {\alpha \over 2}} \right){K_{bX}}m$ ....... (3)

For solvent Y, van't Hoff factor $i = 1 - {{0.7} \over 2} = {{1.3} \over 2}$

$\Delta {T_{bY}} = i{K_{bY}}m \Rightarrow \left( {{{1.3} \over 2}} \right){K_{bY}}m$ ....... (4)

Given that $\Delta {T_{bX}} = 3\Delta {T_{bY}}$. Dividing Eq. (3)/(4), we have

${{\Delta {T_{bX}}} \over {\Delta {T_{bY}}}} = {{\left( {1 - {\alpha \over 2}} \right){K_{bX}}m} \over {\left( {{{1.3} \over 2}} \right){K_{bY}}m}} \Rightarrow {1 \over 3} = {{2(2 - \alpha )} \over {1.3}} \Rightarrow \alpha = 0.05$

It is given that the mole fraction of the solute $\left(x_{\text {solute }}\right)$ is 0.1 , therefore mole fraction of the solvent $\left(x_{\text {solvent }}\right)$ is 0.9 . So we have

$ x_{\text {solute }}=\frac{n_{\text {solute }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.1 $ ......(1)

$ x_{\text {solvent }}=\frac{n_{\text {solvent }}}{\left(n_{\text {solute }}+n_{\text {solvent }}\right)}=0.9 $ ......(2)

where $n_{\text {solute }}$ and $n_{\text {solvent }}$ are the number of moles of solute and solvent, respectively. Dividing Eq. (1) by Eq. (2) gives

$ \frac{x_{\text {solute }}}{x_{\text {solvent }}}=\frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{0.1}{0.9} $ .......(3)

Given that the density of solution is $2 \mathrm{~g} \mathrm{~cm}^{-3}$, we have

$ \begin{aligned} & W_{\text {solution }} =\text { density } \times V_{\text {solution }}=2 \times V_{\text {solution }} \\\\ & \therefore W_{\text {solute }}+W_{\text {solvent }} =2 \times V_{\text {solution }} .......(4) \end{aligned} $

We know that molality is given by

$ m=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times W_{\text {solvent }}} $

Substituting from Eq. (3), we have

$ m=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{W_{\text {solvent }}} $ ......(5)

Molarity is given by

$ M=\frac{n_{\text {solute }} \times 1000}{n_{\text {solvent }} \times V_{\text {solution }}} $

Substituting from Eq. (3), we have

$ M=\left(\frac{0.1}{0.9}\right) \times \frac{1000}{V_{\text {solution }}} $ ..........(6)

Given that, Molarity $(M)=$ Molality $(m)$;

Therefore, from Eqs. (5) and (6), we get

$ \begin{aligned} & \frac{0.1 \times 1000}{0.9 \times W_{\text {solvent }}} =\frac{0.1 \times 1000}{0.9 \times V_{\text {solution }}} \\\\ & W_{\text {solvent }} =V_{\text {solution }} \end{aligned} $

From Eq. (4), we have

$ \begin{aligned} W_{\text {solvent }} & =\frac{W_{\text {solute }}+W_{\text {solvent }}}{2} \\\\ 2 W_{\text {solvent }} & =W_{\text {solute }}+W_{\text {solvent }} \\\\ W_{\text {solvent }} & =W_{\text {solute }} .........(7) \end{aligned} $

The molecular weight (MW) of the solute can be calculated by dividing the weight of the solute by the number of moles of solute. This can be written as :

$ MW_{\text{solute}} = \frac{W_{\text{solute}}}{n_{\text{solute}}} $

2. Similarly, the molecular weight (MW) of the solvent can be calculated by dividing the weight of the solvent by the number of moles of solvent. This can be written as :

$ MW_{\text{solvent}} = \frac{W_{\text{solvent}}}{n_{\text{solvent}}} $

In these formulas, $MW_{\text{solute}}$ and $MW_{\text{solvent}}$ represent the molecular weights of the solute and solvent respectively, $W_{\text{solute}}$ and $W_{\text{solvent}}$ represent their weights, and $n_{\text{solute}}$ and $n_{\text{solvent}}$ represent the number of moles of solute and solvent respectively.

From Eq. (3), we have

$ \frac{n_{\text {solute }}}{n_{\text {solvent }}}=\frac{W_{\text {solute }} / M W_{\text {solute }}}{W_{\text {solvent }} / M W_{\text {solvent }}}=\frac{0.1}{0.9} $

Using Eq. (7), we get

$ \frac{M W_{\text {solute }}}{M W_{\text {solvent }}}=9 $

The depression in freezing point is given by

$\Delta$T_f = K_f $\times$ m $\times$ i

0.0558 = 1.86 $\times$ 0.01 $\times$ i

i = 3

Therefore, one mole of complex gives three moles of ions in solution.

Hence, the complex is [Co(NH₃)₅Cl]Cl₂ and the number of Cl^$-$ ions inside the coordination sphere is 1.

The depression of freezing point is an important colligative property, which is influenced by the number of solute particles in a solution. When a solute dissociates or ionizes in a solution, it increases the number of particles in the solution, thereby affecting colligative properties such as the depression of freezing point. The degree of dissociation ($\alpha$) gives us a measure of the extent to which a compound dissociates into its ions. For our case, MX₂ dissociates into one M²⁺ ion and two X^- ions.

Let's initially consider 1 mole of MX₂ is present. Since the degree of dissociation ($\alpha$) is 0.5, this means half of the MX₂ dissociates into its ions, and half remains undissociated.

For the dissociation reaction:

$\text{MX}_{2} \rightarrow \text{M}^{2+} + 2\text{X}^{-}$

If the initial amount of MX₂ is 1 mole, then:

• The amount of MX₂ that dissociates = $\alpha = 0.5$ moles
• The amount of M²⁺ formed = $\alpha = 0.5$ moles (since for every mole of MX₂ that dissociates, 1 mole of M²⁺ is formed)
• The amount of X^- formed = $2\alpha = 2 \times 0.5 = 1$ mole (since for every mole of MX₂ that dissociates, 2 moles of X^- are formed)

The Van't Hoff factor (i) quantifies the effect of solute particles on the colligative properties of a solution. It is defined as the ratio of the actual number of particles in solution after dissociation to the number of formula units initially dissolved in the solvent.

$i = \frac{\text{Total number of particles after dissociation}}{\text{Number of moles of solute originally dissolved}}$

Considering the dissociation and the amounts calculated:

• Total number of particles after dissociation = (undissociated MX₂) + (M²⁺) + (X^-) = $(1 - \alpha) + \alpha + 2\alpha$
• Putting the value of $\alpha = 0.5$, we get:

  $i = \frac{(1 - 0.5) + 0.5 + 2(0.5)}{1} = \frac{1 + 1}{1} = 2$

Now, the depression in freezing point ($\Delta T_f$) is directly proportional to the molal concentration of the solute particles and Van't Hoff factor (i):

$\Delta T_f = i \cdot K_f \cdot m$

Where $K_f$ is the cryoscopic constant and $m$ is the molality of the solution.

Without ionic dissociation, the value of $i$ would have been 1 (since the solute would not have dissociated into multiple particles). However, due to the dissociation, the value of $i$ has increased to 2. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation, therefore, would be directly equal to the ratio of the Van't Hoff factors:

$\frac{\text{Observed } \Delta T_f}{\text{In the absence of ionic dissociation}} = \frac{2 \cdot K_f \cdot m}{1 \cdot K_f \cdot m} = \frac{2}{1} = 2$

This ratio shows that due to the ionic dissociation of MX₂ into M²⁺ and X^- ions with a degree of dissociation ($\alpha$) of 0.5, the observed depression in freezing point is twice the value it would have been in the absence of ionic dissociation.

Step 1: Calculate moles of phenol.
The number of moles of phenol is ${{75.2} \over {94}} = 0.8$ mol.
(Here, the molar mass of phenol is 94 g/mol, and 75.2 g is dissolved.)

Step 2: Find the molality of the solution.
Molality (m) = ${{0.8} \over {1}} = 0.8$ mol/kg.
(Because 0.8 mol phenol is dissolved in 1 kg of solvent.)

Step 3: Use the formula for depression in freezing point.
We know $\Delta T_f = k_f \times m$
Given: $k_f = 14$, so substitute the values.

$\Delta T_f = 14 \times 0.8 = 11.2~\mathrm{K}$
This is the calculated value if no dimerization happened.

Step 4: Compare with the experimental value.
The actual (experimental) depression in freezing point is 7 K, which is less than 11.2 K.

Step 5: Find the van't Hoff factor (i).
$i = {7 \over 11.2} = 0.625$
(This value is less than 1. It shows association, not dissociation.)

Step 6: Find degree of association (α).
The formula is: $\alpha = {{i - 1} \over {{1 \over n} - 1}}$
where n = number of molecules coming together. For dimerization, n = 2.

Substitute values: $\alpha = {{0.625 - 1} \over {{1 \over 2} - 1}} = {{-0.375} \over {-0.5}} = 0.75$

Step 7: Calculate the percent dimerization.
The percentage is $0.75 \times 100 = 75\%$.

Answer: 75% of the phenol molecules dimerize in the solution.