Liquid Solution

$ \alpha=0.5 $ (50\% dissociated)

The reaction involves is

$ \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right] \rightleftharpoons 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4} $

$\begin{array}{1} \text { Initial moles } & 1 & 0 & 0 \\ \text { Moles after dissociation } & 1-\alpha & 4 \alpha & \alpha \end{array}$

Total moles at equilibrium $ =1-\alpha+4 \alpha+\alpha $

Total $ =1+4 \alpha $

$ \therefore $ van't Hoff factor $ (i)=\frac{1+4 \alpha}{1}=1+4 \times 0.5=3 $

Osmotic pressure, $ \pi=i C R T $

$ \begin{aligned} C & =0.1 \mathrm{M}=10^{2} \mathrm{~mol} \mathrm{~m}^{-3} \\ \pi & =3 \times 10^{2} \times 8.314 \times 300 \\ & =7.483 \times 10^{5} \mathrm{Nm}^{-2} \end{aligned} $

or $ \quad \pi=7.48 \mathrm{~atm} $

$ 58.5 \mathrm{~g} \mathrm{NaCl}=\frac{58.5}{58.5} \mathrm{~mol} $.

$ \begin{aligned} & =1 \mathrm{~mol} \\ 180 \mathrm{~g} \text { of glucose } & =\frac{180}{180}=1 \mathrm{~mol} \end{aligned} $

Thus, we have 1 M NaCl and 1 M glucose solution.

But NaCl dissociates in the solution and glucose does not dissociates.

Thus particle concentration is more in NaCl .

Hence, NaCl solution will have higher elevation in boiling point.

We know that,

elevation in boiling point is given by

$ \begin{aligned} \Delta T_b & =K_b \times \text { molality } \\ \mathrm{[}\Delta T_b & =T_{\text {solute }}-T_{\text {solvent }}^{\circ} \mathrm{]}=100.17^{\circ} \mathrm{C}-100^{\circ} \mathrm{C} \\ & =0.17^{\circ} \mathrm{C} \end{aligned} $

$0.17^{\circ} \mathrm{C}=0.512 \mathrm{k} \mathrm{kg} \mathrm{mol}^{-1} \times$ Molality

$ \text { Molality }=\frac{0.17}{0.512} \mathrm{~m} $

Let depression in freezing point be $\Delta T_f$.

$ \begin{aligned} \Delta T_f & =\text { Molality } \times K_f=\frac{0.17}{0.512} \times 1.86 \\ & =0.62^{\circ} \mathrm{C} \end{aligned} $

Thus, freezing point of solution

$ \begin{aligned} & =T_{\text {Solvent }}^0-\Delta T_f=0^{\circ} \mathrm{C}-0.62 \\ & =-0.62^{\circ} \mathrm{C} \end{aligned} $

$p^{\circ}=268$ torr $p_s=167$ torr $\left[\frac{p^{\circ}-p_s}{p^{\circ}}=\chi_B\right]$ mole fraction of solute

$ \chi_B=\frac{n_B}{n_B+1}=0.3768 $

or, $\quad 0.3768 n_B+0.3768=n_B$

or $\quad n_B=0.605$

Given, the vapour pressure of

$ p_A^{\circ}=50 \mathrm{~mm} \mathrm{Hg} $

The vapour pressure of

$ \begin{aligned} p_B^{\circ} & =32 \mathrm{~mm} \mathrm{Hg} \\ \text { Temperature } & =300 \mathrm{~K} \\ n_A & =n_B=1 \mathrm{~mole} \end{aligned} $

$ \begin{aligned} \therefore \quad X_A & =\frac{n_A}{n_A+n_B}=\frac{1}{1+1}=0.5 \\ X_A+X_B & =1 \Rightarrow X_B=1-0.5=0.5 \end{aligned} $

According to Raoult's law,

$ \begin{aligned}For A, \,\,p_A=p_A^{\circ} \times X_A & =50 \times 0.5 \\ & =25 \mathrm{~mm} \mathrm{Hg} \end{aligned} $

$ \begin{aligned}For B,\,\, p_B & =p_B^{\circ} \times X_B \\ & =32 \times 0.5=16 \mathrm{~mm} \mathrm{Hg} \\ p_T & =p_A+p_B=25+16 \\ & =41 \mathrm{~mm} \mathrm{Hg} \end{aligned} $

Mole fraction of $A$ in vapour phase,

$ Y_A=\frac{25}{41}=0.609 \simeq 0.61 $

Elevation in boiling point $\Delta T_b=$ Boiling point of solution - Boiling point of pure solvent

$ \begin{aligned} & =373.202 \mathrm{~K}-373.15 \mathrm{~K}=0.052 \mathrm{~K} \\ & \text { Also, } \Delta T_b=K_b \times \text { Molarity } \\ & \Delta T_b=K_b \times \end{aligned} $

$\frac{\text { Number of moles of solute (urea) }}{\text { Mass of solvent (in kg) }}$

$ \begin{aligned} & 0.052=0.52 \times \frac{\frac{x}{60}}{\frac{y}{1000}} \\ & 0.052=0.52 \times \frac{x}{60 \times y} \times 1000 \\ & \frac{0.052 \times 60}{0.52 \times 1000}=\frac{x}{y} \\ & 6 \times 10^{-3}=x: y \end{aligned} $

To determine the degree of dissociation ($\alpha$) of sodium chloride at 300 K given the osmotic pressure, we can use the van't Hoff factor ($i$) and the formula for osmotic pressure ($\pi$).

Given

Osmotic pressure ($\pi$) = 4.82 atm

Temperature ($T$) = 300 K

Concentration ($C$) = 0.1 N

$R$ (gas constant) = 0.082 L atm K$^{-1}$ mol$^{-1}$

Formula for Osmotic Pressure

$ \pi = i C R T $

Calculation

First, calculate the van't Hoff factor ($i$):

$ i = \frac{\pi}{C R T} = \frac{4.82}{0.1 \times 0.082 \times 300} = 1.96 $

Understanding the Dissociation

For sodium chloride ($\text{NaCl}$), which dissociates into two ions $\text{Na}^+$ and $\text{Cl}^-$:

$ i = 1 + \alpha $

where $n$ (number of ions) for NaCl is 2. Therefore:

$ 1.96 = 1 + \alpha $

Solving for $\alpha$:

$ \alpha = 1.96 - 1 = 0.96 $

Thus, the degree of dissociation $\alpha = 96 \times 10^{-2}$.

Conclusion

Therefore, $x = 96$.

Given:

Vapour pressure of pure liquid $ A $, $ p_A^{\circ} = 70 \text{ torr} $

Mole fraction of liquid $ B $, $ \chi_B = 0.2 $

Total vapour pressure of the solution, $ p_{\text{total}} = 84 \text{ torr} $

Vapour pressure of pure liquid $ B $, $ p_B^{\circ} = ? $

First, identify the mole fraction of liquid $ A $:

$ \chi_A + \chi_B = 1 $

$ \chi_A = 1 - 0.2 = 0.8 $

Using Dalton's law of partial pressures, express the total vapor pressure:

$ p_{\text{total}} = p_A + p_B $

Apply Raoult's law to determine the partial pressures of components $ A $ and $ B $:

For component $ A $:

$ p_A = p_A^{\circ} \cdot \chi_A $

$ p_A = 70 \times 0.8 = 56 \text{ torr} $

For component $ B $:

$ p_B = p_B^{\circ} \cdot \chi_B = p_B^{\circ} \times 0.2 $

Substitute the values found for $ p_A $ and the expression for $ p_B $ into the equation for total pressure:

$ 84 = 56 + 0.2 \times p_B^{\circ} $

Solve for $ p_B^{\circ} $:

$ p_B^{\circ} = \frac{84 - 56}{0.2} = 140 \text{ torr} $

$w=2 \mathrm{~g}, M$ of $\mathrm{NaOH}=40$

$V$ of solution $=200 \mathrm{~mL}$

$ \begin{aligned} \text { Molarity } & =\frac{w}{M} \times \frac{1000}{V(\mathrm{~mL})} \\ & =\frac{2}{40} \times \frac{1000}{200}=0.25 \text { at } 25^{\circ} \mathrm{C} \end{aligned} $

But as temperature increases, water expands,

∴ The volume of solution increases.

Hence at $90^{\circ} \mathrm{C}$ the molarity will be less than 0.25 .

$ \begin{aligned} &\text { }\\ &\begin{aligned} K_f & =? T=17^{\circ} \mathrm{C}=17+273=290 \mathrm{~K} \\ K_f & =\frac{R T_f^2}{1000 \times L_f}=\frac{8.314 \times(290)^2}{1000 \times 180} \\ & =3.88 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} . \end{aligned} \end{aligned} $

A liquid mixture obeys Raoult's law at all concentrations in which solute-solute, solute-solvent and solvent-solvent interactions are similar.

Freezing point

$ \propto \frac{1}{\text { van't Hoff factor }(i)} $

For $\mathrm{C}_6 \mathrm{H}_5-\stackrel{+}{\mathrm{N}} \mathrm{HNH}_3 \mathrm{Cl}^{-}, i=2$

For $\mathrm{Ca}\left(\mathrm{NO}_3\right)_2, i=3$

For $\mathrm{LaCl}_3, i=4$

For $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6, i=1$

So, $\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6$ has minimum value of van't Hoff factor, hence, its freezing point will be highest.

According to Henry's law, $p=K_{\mathrm{H}} \times x$

$ \begin{aligned} p & =\text { partial pressure of gas } \\ x & =\text { mole fraction of gas in solution } \\ K & =\text { Henry's constant } \\ p & =K_{\mathrm{H}} \times x \\ x & =\frac{2 \times 1.013 \times 10^5}{1.67 \times 10^8} \end{aligned} $

$ \begin{array}{ll} & \frac{n_{\mathrm{CO}_2}}{1000 / 18}=\frac{2 \times 1.013}{1.67 \times 1000} \\ \Rightarrow & n_{\mathrm{CO}_2}=\frac{2.026 \times 1000}{1.67 \times 1000 \times 18} \\ \Rightarrow & \frac{\text { mass of } \mathrm{CO}_2}{44}= \\ \text { Mass of } \mathrm{CO}_2 \Rightarrow 2.96 \mathrm{~g} \end{array} $

The highest is the number of particles in the solution, the highest will be the colligative property. $0.1 \mathrm{M}\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ will produce highest number of ions in the solution, i.e. $3 \mathrm{NH}_4^{+}$ions and one $\mathrm{PO}_4^{3-}$ ion. So, $\left(\mathrm{NH}_4\right)_3 \mathrm{PO}_4$ will show the highest colligative property.

Given, $K_H=10^5 \mathrm{~atm}$

Pressure, $p=5 \mathrm{~atm}$

Mole fraction of $\mathrm{N}_2$ in air, $\chi_{\mathrm{N}_2}=0.8$

Partial pressure of $\mathrm{N}_2$,

$ \begin{aligned} p_{\mathrm{N}_2} & =\chi_{\mathrm{N}_2} \times p \\ & =0.8 \times 5=4 \mathrm{~atm} \end{aligned} $

According to Henry's law,

$ \begin{aligned} & p_{\mathrm{N}_2}=K_H \chi_{\mathrm{N}_2}^{\prime} \\ & \chi_{\mathrm{N}_2}^{\prime}=\frac{4}{10^5}=4 \times 10^{-5} \end{aligned} $

It means, 1 mole water contains $4 \times 10^{-5}$ moles of $\mathrm{N}_2$.

$\therefore 10$ moles of water contains

$ \begin{aligned} & =4 \times 10^{-5} \times 10 \\ & =4 \times 10^{-4} \text { moles of } N_2 \end{aligned} $

Osmotic pressure is a colligative property. So, it depends only on the concentration or amount of solute and temperature.

So, osmotic pressure of a solution remains unchanged on increasing or decreasing the external pressure.

In $\mathrm{CH}_3 \mathrm{OH}$ and $\mathrm{CH}_3 \mathrm{COOH}$ mixture solution, $A-A$ and $B-B$ interactions are weaker than

$A-B$ interactions. It leads to increase in vapour pressure and positive deviation Raoult's law. For solutions showing positive deviation from Raoult's law, $\Delta H_{\text {mix }}>0$ and $\Delta V_{\text {mix }}>0$. Thus, 1 and 4 are incorrect. According to which the correct option is (a).

Henry's law states that, at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas present above the surface of liquid or solution. It is applicable only to non-reacting mixtures of liquid and gas.

On dissolution, ammonia gas reacts with water to form ammonium hydroxide. Oxygen gas dissolves in unsaturated blood to form oxyhaemoglobin. Therefore, they do not follow Henry's law.

$\mathrm{O}_2$ gas and $\mathrm{CO}_2$ gas do not react with water on dissolution. Therefore, they follow Henry's law.

In ideal solutions, the volume of the solution is the sum of the volumes of the components before mixing.

$ V_{\text {solvent }}+V_{\text {solute }}=V_{\text {solution }} $

i.e. there is no change in volume on mixing or $\Delta V_{\text {mix }}$ is zero.

The intermolecular interactions between the components ( $A-B$ interactions) are of same magnitude as the intermolecular interactions in pure components ( $A-A$ and $B$ - $B$ interactions). Since, there is no change in magnitude of the intermolecular interactions in the two components present, the heat change on mixing, i.e. $\Delta H_{\text {mix }}$ in such solutions must be zero.

$ \mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \longrightarrow \mathrm{H}_2 \mathrm{CO}_3{ }^{-} $

This is a reaction in which carbonic acid is formed from water and $\mathrm{CO}_2$. So, this is not an ideal solution.

We have the equation $\pi=i C R T$. As urea is non-electrolyte, $i=1, \pi$ is the osmotic pressure at temperature $T$.

$ \begin{aligned} \pi_1 & =C_1 R T_1 \\ 400 & =C_1 R \times 273 \mathrm{~K} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \pi_2 & =C_2 R T_2 \\ 100 & =C_2 R \times 293 \mathrm{~K} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Dividing Eq. (i) by Eq. (ii),

$ \begin{aligned} & \frac{\pi_1}{\pi_2}=\frac{400}{100}=\frac{C_1 \times 273}{C_2 \times 293} \\ & \frac{C_1}{C_2}=\frac{400 \times 293}{100 \times 273}=4.29 \end{aligned} $

$C_1$ is 4.29 times $C_2$. So, the dilution factor is 4.3 .

$\begin{aligned} \text { Given, } n & =0.05 \mathrm{~mol} \\ V & =500 \mathrm{~g} \text { of water } \\ K_f & =1.86 \mathrm{~k} \mathrm{~kg} / \mathrm{mol} \\ \Delta T_f & =? \end{aligned}$

$\begin{aligned} \therefore \text { Molality } & =\frac{\text { Number of moles }}{\text { Weight of solvent }(\mathrm{kg})} \\ & =\frac{0.05}{0.5} \\ \Delta T_f & =m \times K_f \\ & =\frac{0.05}{0.5} \times 1.86 \\ & =0.186 \mathrm{~K} \end{aligned}$

Ideal solutions are the solutions which obey Raoult's law at all temperatures and pressures. The interactions of solute-solute, solvent-solvent and solute-solvent in these solutions are almost similar. No association/dissociation occurs in these solutions.

Mixture of phenol and aniline shows negative deviation from ideal behaviour and forms an azeotropic mixture. The phenol-aniline interactions are stronger than phenol-phenol or aniline-aniline interactions.

$\begin{aligned} & \Delta V_{\text {mixing }}<0 \\ & \Delta H_{\text {mixing }}>0 \end{aligned}$

Hence, it does not form an ideal solution.

Rest options (I), (II) and (III) form ideal solution.

Mass of urea, $m_U=6 \mathrm{~g}$

Molar mass of urea, $M_U=60 \mathrm{~g} \mathrm{~mol}^{-1}$

Moles of urea, $n_v=6 / 60=0.1 \mathrm{~mol}$

Mass of $X, m_X=10 \mathrm{~g}$

As solutions are isotonic, so

$\begin{aligned} & \pi_U=\pi_x \\ & C_U R T=C_x R T \end{aligned}$

Assuming volume of solvent to be same in both cases,

$ \begin{aligned} & n_v=n_X \\ & 0.1=n_X \quad \text{where, } m_x= \text{mass of X} \end{aligned}$

$\Rightarrow \quad \frac{m_X}{M_X}=0.1 \quad M_X=\text { Molar mass of } X$

$M_X=\frac{10}{0.1}=100 \mathrm{~g} \mathrm{~mol}^{-1}$

$4 \% w / V$ solution means 100 mL of solution contains 4 g of non-volatile solute.

Molar mass of solute $=120 \mathrm{~g} \mathrm{~mol}^{-1}$

Moles, $n_s=\frac{4}{120}=\frac{1}{30} \mathrm{~mol}$

$x \%\left(\frac{w}{V}\right)$, so mass of urea is $x \mathrm{~g}$.

Molar mass of urea $=60 \mathrm{~g} \mathrm{~mol}^{-1}$

Using formula, number of moles $=\frac{\text { Mass }}{\text { Molar mass }}$

Moles, $n_U=\frac{x}{60} \mathrm{~mol}$

For isotonic solution, $\pi_s=\pi_U$

$\begin{aligned} C_o & =C_S \\ n_U & =n_S \\ \frac{x}{60} & =\frac{1}{30} \\ x & =2 \end{aligned}$

$\therefore 2 \%(w / V)$ solution of urea is isotonic with $4 \% ~(w / V)$ solution of non-volatile solute.

Given, mass of non-volatile solid = x g

Molar mass = 78 g mol$^{-1}$

Mass of water = 0.5 kg

$K_f \text { (water) }=1.86 \mathrm{~K} \mathrm{~kg}^{-1} \mathrm{~mol}^{-1}$

Lowering in freezing point, $\Delta T_f=1^{\circ} \mathrm{C}$ Using the equation;

$\Delta T_f=K_f m \Rightarrow m=\frac{\Delta T_f}{K_f}$

Now, substituting the given values in above formula,

$m=\frac{1}{1.86} \mathrm{~kg}$

We know that molality

$\begin{aligned} &= \frac{\text { Number of moles of solute }}{\text { Mass of solvent (in } \mathrm{kg})} \\ & \frac{n}{M}=\frac{1}{1.86} ; \quad[n=\text { number of moles of solute } \\ & \text { So, } n=\left.\frac{1}{1.86} \times 0.5=\text { mass of solvent in } \mathrm{kg}\right] \\ & \frac{x}{78}=\frac{0.5}{1.86} \\ & \Rightarrow \quad x=\frac{0.5 \times 78}{1.86} \\ & x=20.96 \mathrm{~g} \end{aligned}$

Blood serum contains $0.9 \%$ salt. When blood cells placed in saline water, they lose water due to exosmosis and collapses. Thus, assertion is correct.

The water moves through cell membrane during this process. Thus, reason is not correct.

Hence, option (c) is correct. i.e. A is correct but R is not correct.

When a minimum boiling azeotrope is formed, the resulting solution will have maximum vapour pressure as it has minimum boiling point.

Interaction between $A-A$ or $B-B$ are stronger than $A-B$ interaction.

$\therefore \Delta V > 0, \Delta H_{\text {mix }} > 0 $

$\therefore$ Only statement $(\mathrm{b})$ is correct.

The decrease in vapour pressure can be calculated as:

$p^\Upsilon - p_s = p^\Upsilon \times \text{mole fraction of solute}$

We are told that the vapour pressure goes down by 20 mm of Hg when the mole fraction of solute is 0.5. So,

$20 = p^\Upsilon \times 0.5$     ... (i)

If we want the vapour pressure to drop by only 10 mm of Hg, let the new mole fraction of solute be $\chi_2$:

$10 = p^\Upsilon \times \chi_2$     ... (ii)

Now, let's solve for $\chi_2$:

From equation (i): $p^\Upsilon = 40$

Using this in equation (ii):

$10 = 40 \times \chi_2$

So, $\chi_2 = \frac{10}{40} = \frac{1}{4}$

The mole fraction of the solvent ($\chi_1$) is what is left after subtracting the solute's mole fraction from 1:

$\chi_1 = 1 - \frac{1}{4} = \frac{3}{4}$

Higher the value of K$_H$, lower will be the solubility of the gas at a given pressure.

Solubility $\propto$ $\frac{1}{K_H}$

So, the correct order is

Ar < CO$_2$ < CH$_4$ < HCHO

$\begin{aligned} & \mathrm{CaCl}_2 \longrightarrow \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \\ & \mathrm{Cl}^{-}=3.01 \times 10^{22} \text { ions } \end{aligned}$

For

$\begin{aligned} 2 \mathrm{Cl}^{-} & =\frac{3.01 \times 10^{22}}{2} \\ & =1.505 \times 10^{22} \mathrm{CaCl}_2 \text { required } \\ \text { Mole } & =\frac{1.505 \times 10^{22}}{6.02 \times 10^{23}}=\frac{0.25}{10}=0.025 \mathrm{~mol} \end{aligned}$

Molarity for 0.025 mole of $\mathrm{CaCl}_2$

$=\frac{0.025}{500} \times 1000=0.05 \mathrm{M} $

An ideal solution is a homogeneous mixture which $\left(\Delta_{\text {mix }} S=0\right)$ obey Raoult's law.

For ideal solution, we need $\Delta_{\text {mix }} H=0$ and $\Delta_{\text {mix }} V=0$ but $\Delta_{\text {mix }} S \neq 0$ i.e. $\Delta_{\text {mix }} S > 0$ obey Raoult's law at every range of concentration for $\Delta_{\text {mix }} S > 0$.

Number of molecule $=$ entropy increase

$\begin{aligned} & \Delta_{\text {mix }} G=\Delta H-T \Delta S \\ & \Delta_{\text {mix }} G=-T \Delta S<0 \end{aligned}$

e.g., benzene + toluene (show ideal solution)

Formula used,

Freezing point of solution, $\Delta T_l=K_f \times$ molality of solution and $\Delta T_b=K_b \times$ molatity of solution

Now, ratio of $K_f$ and $K_b$ is $\frac{\Delta T_f}{\Delta T_b}=\frac{K_f}{K_b}$ ..... (i)

Given that, $\Delta T_b=T_2-T_1=100.20-100=0.20^{\circ} \mathrm{C}$

$K_f$ of water $=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

$K_b$ of water $=0.512 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

Put value in Eq. (i),

$\therefore \quad \frac{\Delta T_f}{0.20}=\frac{1.86}{0.512} \Rightarrow \Delta T_f=0.726$

Now, $\Delta T_f \Rightarrow T_1-T_2$

$0.726^{\circ} \mathrm{C}=0^{\circ} \mathrm{C}-T_2 \Rightarrow T_2=-0.726^{\circ} \mathrm{C}$

Hence, freezing point of solution is $-0.726^{\circ} \mathrm{C}$

Fractional distillation is used for the separation of a mixture of two or more miscible liquids for which the difference in boiling points is less than 25K.

e.g. CCl$_4$ (Carbon tetrachloride) boiling point is 350 K and anisole (boiling point is 427 K can be separated by fractional distillation.

The process involves heating the mixture and partial condensation of vapours along a column, which is set up such that components with lower boiling points pass through the column and are collected earlier than components with higher boiling points.

Solutions of benzene and toluene, which have very similar molecular structure, are ideal. Any mixture of the two has a volume equal to the sum of the volumes of the separate components and the mixing process occurs without absorption or evolution of heat.

Benzene and toluene solution over the entire range of composition is ideal. The vapour pressure of pure benzene and toluene at 300 K are 50.

The molal elevation constant is the ratio of elevation in boiling point to molality of solution.

$\begin{aligned} \Delta T_b & =K_b \times m \\ K_b & =\frac{\Delta T_b}{m} \end{aligned}$

Here, $K_b=$ molal elevation constant

$\Delta T_b$ = increase in boiling point

$m$ = molality of solution

Due to solute presence in solution, the solution's boiling point is greater than the pure solvent's boiling point. This is called elevation in boiling point $\left(\Delta T_b\right)$.

According to Raoult's law, relative lowering of vapour pressure is equal to mole fraction of solute,

$\therefore$ ${{p - {p_s}} \over p} = {n \over {n + N}} = {w \over m}{{w/m} \over {{w \over m} + {W \over M}}} \Rightarrow {w \over m} < < < {W \over M}$

$\therefore$ ${{p^\circ - {p_s}} \over {p^\circ }} = {{w/m} \over {{W \over M}}} = {w \over m} \times {M \over W}$

$0.0125 = {{w \times 18} \over {m \times W}} \Rightarrow {w \over {mW}} = {{0.0125} \over {18}}$

$ = {{Weight\,of\,solute\,(w) \times 1000} \over {Molar\,mass\,of\,solute\,(m) \times Weight\,of\,{H_2}O(W)}}$

Now, molality $m = {{0.0125} \over {18}} \times 1000 = 0.69 \approx 0.7$

$\because$ Molar mass of glucose $(M)=180 \mathrm{~g}$

Molar mass of solute $(A)$,

i.e. $\left(M_A\right)=58.5$

\% of $A=1.17 \%$

\% of glucose = 7.2\%

Also,

$\because \pi($ osmotic pressure $)=C R T$

$ \pi=\frac{n}{V} \cdot R T \Rightarrow \pi=\frac{W}{M} \times R T $

where, $\pi=$ osmotic pressure, $w=\%$ of sample

$ \begin{aligned} M & =\text { molar mass, } R=\text { gas constant } \\ T & =\text { temperature } \end{aligned} $

∵ These are isotonic.

$ \begin{aligned} \therefore \quad & i=\frac{M_A \times \% \text { of glucose }}{\% \text { of } A \times M} \\ & i=\frac{58.5 \times 7.2}{1.17 \times 180}=\frac{421.2}{210.6}=2 \end{aligned} $

Hence, option (a) is the correct answer.

$\mathrm{Na}_2 \mathrm{O}$ and $\mathrm{KO}_2$, both are ionic compounds and are completely ionised as follows :

(i) $\mathrm{Na}_2 \mathrm{O} \longrightarrow 2 \mathrm{Na}^{+}+\mathrm{O}^{2-}(i=3)$

(ii) $\mathrm{KO}_2 \longrightarrow \mathrm{~K}^{+}+2 \mathrm{O}^{2-}(i=3)$

Also, $\because \frac{p^{\circ}-p_s}{p^{\circ}}=i \cdot \chi_{\mathrm{Na}_2 \mathrm{O}}+i \cdot \chi_{\mathrm{KO}_2}$

and moles of $\mathrm{H}_2 \mathrm{O}=\frac{1000}{18}=55.5$

where, $p^{\circ}=$ atmospheric pressure

$p_s=$ vapour pressure of solution

$p^{\circ}=760$

$\therefore \frac{p^{\circ}-p_s}{p^{\circ}}=3 \times \frac{3}{3+3+55.5}+3$\times \frac{1.5}{1.5+1.5+55.5}$

Hence, option (d) is the correct answer.

Relative lowering of vapour pressure of a non-volatile solute is expressed as :

$ \mathrm{RLVP}=\frac{\Delta p}{p^{\circ}}=\chi_B=0.5 \text { (given) } $

where,

$ \begin{aligned} \chi_B & =\text { mole fraction of the solute } \\ & =0.5 \end{aligned} $

Let, solubility of $A B_2$ ( $1: 2$ type electrolyte) is pure water

$ \begin{aligned} & =5 \mathrm{~mol} \mathrm{~L}^{-1}=5 \mathrm{M} \\ \Rightarrow \quad K_{\mathrm{sp}} & =4 S^3 \\ & =2.56 \times 10^{-4} \mathrm{M}^3 \text { (given) } \\ \therefore \quad S & =0.04 \mathrm{M} \end{aligned} $

$=0.04 \mathrm{~m}$ (molal) under standard condition of the solution.

Depression of freezing point,

$ \begin{aligned} \Delta T_f & =K_f \times m \times i \\ & =1.8 \times 0.04 \times 3 \\ & =0.216 \mathrm{~K} \end{aligned} $

[ ∵ Assuming complete dissociation of $A B_2$, van't Hoff factor, $i=3$ ]

$ \begin{aligned} & \because \text { Volume }=\frac{\text { Mass }(m)}{\text { Density }(d)}=\frac{100}{1.02}=98.039 \mathrm{~mL} \\ & \therefore \quad \text { Molarity }=\frac{9.8 \times 1000}{98 \times 98.04}=1.01 \end{aligned} $

Which is close to 1.1 .

Hence, option (a) is the correct answer.

The freezing point among the following equimolal aqueous solutions will be highest for urea because urea does not dissociate. So, it has the minimum number of particles and

therefore, it shows minimum depression in freezing point. So, it has the maximum freezing point.

$\therefore$ ${K_b} = {{RT_b^2{M_1}} \over {\Delta {H_{vap}}}} = {{8.314 \times {{(373)}^2} \times 0.0180} \over {40585}}$ = 0.513 K kg/mol

Now, molecular weight of $A{l_2}{(S{O_4})_3} = 342$

$\mathop {A{l_2}{{(S{O_4})}_3}}\limits_{1 - \alpha } $ $\rightleftharpoons$ $\mathop {2A{l^{3 + }}}\limits_{2\alpha } $ + $\mathop {3SO_2^{2 - }}\limits_{2\alpha } $

i = 1 $-$ $\alpha$

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

$ = 5 \times 0.513 \times {{5.6} \over {342 \times 1}} = 0.042$

Boiling point of solution = 100 + 0.042 = 100.042$^\circ$C