Liquid Solution

The problem states that 1 mL of the $\mathrm{H}_2 \mathrm{O}_2$ solution yields 20 mL of oxygen gas at STP. By definition, " $x$ volume" $\mathrm{H}_2 \mathrm{O}_2$ means that 1 mL of the solution produces $x \mathrm{~mL}$ of $\mathrm{O}_2$ at STP. Therefore, the volume strength, $x$, is 20 .

The decomposition of hydrogen peroxide is:

$ 2 \mathrm{H}_2 \mathrm{O}_2 \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 $

From the stoichiometry, 2 moles of $\mathrm{H}_2 \mathrm{O}_2 (2 \times 34 \mathrm{~g}=68 \mathrm{~g})$ produce 1 mole of $\mathrm{O}_2 (22.4 \mathrm{~L}$ at STP).

If an $\mathrm{H}_2 \mathrm{O}_2$ solution has a volume strength of ' $x$ ', it means $1 \mathrm{~L}(1000 \mathrm{~mL})$ of the solution produces ' $x$ ' L of $\mathrm{O}_2$ at STP.

Mass of $\mathrm{H}_2 \mathrm{O}_2$ required to produce ' $x$ ' L of $\mathrm{O}_2$.

$ \text {} \begin{aligned} Mass \,\,of\,\, \mathrm{H}_2 \mathrm{O}_2 & =\left(\frac{68 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2}{22.4 \mathrm{~L} \mathrm{O}_2}\right) \times\left(x \mathrm{~L} \mathrm{O}_2\right) \\ & =\frac{68 x}{22.4} \mathrm{~g} \end{aligned} $

This mass of $\mathrm{H}_2 \mathrm{O}_2$ is present in 1000 mL of the solution.

The $(w / v) \%$ is calculated as

$ \begin{aligned} & (w / v) \%=\frac{\text { Mass of solute }(\mathrm{g})}{\text { Volume of solution }(\mathrm{mL})} \times 100 \\ & (w / v) \%=\frac{\frac{68 x}{22.4}}{1000} \times 100 \\ & (w / v) \%=\frac{68 x}{22.4 \times 10}=\frac{68 x}{224}=\frac{17 x}{56} \end{aligned} $

By substituting $x=20,(w / V) \%=6.06 \%$.

Dalton's Law:

The fraction of vapour pressure from $A$ in the total vapour is equal to its mole fraction in the vapour.

$ \frac{p_A}{p_{\text{Total}}} = x_1 $

So, $ p_A = x_1 p_{\text{Total}} \qquad ...(i) $

Raoult's Law:

The vapour pressure of $A$ above the liquid equals its mole fraction in the liquid multiplied by its pure vapour pressure.

$ p_A = p_A^{\circ} x_2 \qquad ...(ii) $

Combine Both Laws:

From equation (i) and (ii), set $p_A$ equal:

$ x_1 p_{\text{Total}} = p_A^{\circ} x_2 $

Solve for the total vapour pressure:

$ p_{\text{Total}} = \frac{p_A^{\circ} x_2}{x_1} $

We use Raoult's Law,

$ p_{\text {total }}=p_A^0 X_A+p_B^0 X_B . $

For the first solution ( 1 mole A, 3 moles B, total 4 moles), $p_{\text {total }}=550 \mathrm{~mm} \mathrm{Hg}$.

$ \begin{aligned} 550 & =p_A^0(1 / 4)+p_B^0(3 / 4) \\ 2200 & =p_A^0+3 p_B^0 \end{aligned} $

For the second solution (1 mole A, 4 moles B , total 5 moles, after adding 1 more mole of B ),

$ \begin{aligned} p_{\text {total }} & =560 \mathrm{~mm} \mathrm{Hg} . \\ 560 & =p_A^0(\mathrm{l} / \mathrm{S})+p_B^0(4 / \mathrm{S}) \\ 2800 & =p_A^0+4 p_B^0 \end{aligned} $

Subtracting Equation 1 from Equation 2

$ \left(p_A^0+4 p_B^0\right)-\left(p_A^0+3 p_B^0\right)=2800-2200 $

This simplifies to $p_B^0=600 \mathrm{~mm} \mathrm{Hg}$ Substituting $p_B^0=600 \mathrm{~mm} \mathrm{Hg}$ into equation 1

$ \begin{aligned} & p_A^0+3(600)=2200 \\ & p_A^0+1800=2200 \\ & p_A^0=400 \mathrm{~mm} \mathrm{Hg} \end{aligned} $

Now, the ratio of $p_A^{\circ}$ and $p_B^{\circ}$ is $\frac{400}{600}=\frac{2}{3}$

Aniline and $\mathrm{H}_2 \mathrm{O}$ can be separated by using steam distillation. They have different boiling point, thus can be seperated by using vapour formation.

Using Henry's law,

Solubility of gas in liquid $\propto \frac{1}{K_{\mathrm{H}}}$

Lower the value of $K_{\mathrm{H}}$, higher will be the solubility.

Thus, the correct order of solubility is

$ \mathrm{HCHO}>\mathrm{CH}_4>\mathrm{CO}_2>\mathrm{Ar} $

$ \mathrm{NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-}$

$ \begin{aligned} & i_{\mathrm{NaCl}}=1+\alpha(n-1) \\ & n=2 ; i_{\mathrm{NaCl}}=1+0.94(2-1)=1.94 \\ & i_{\text {glucose }}=1 \end{aligned} $

Concentration of $\mathrm{NaCl}=\frac{0.01}{0.5}=0.02 \mathrm{M}$

Effective concentration of $\mathrm{NaCl}=i C_{\mathrm{NaCl}}$

$ \begin{aligned} & =1.94 \times 0.02 \mathrm{M}=0.0388 \mathrm{M} \\ C_{\text {glucose }} & =\frac{0.03}{0.5}=0.06 \mathrm{M} \end{aligned} $

Effective concentration $=0.06 \mathrm{M}$

$ \begin{aligned} i C_{\text {total }} & =0.0388+0.06=0.0988 \mathrm{M} \\ \pi & =i C_{\text {total }} R T \\ & =0.098 \times 0.082 \times 300.15=2.43 \mathrm{~atm} \end{aligned} $

Mass of solute $=$ Molarity $\times$ Volume $(\mathrm{L}) \times$ Molar mass

(i) $0.25 \times 1 \times 106=26.5 \mathrm{~g}$

(ii) $0.2 \times 0.25 \times 142=7.1 \mathrm{~g}$

(iii) $1 \times 0.5 \times 158=79 \mathrm{~g}$

(iv) $0.5 \times 0.75 \times 60=22.5 \mathrm{~g}$

Hence, 0.5 L of $1.0 \mathrm{M} \mathrm{KMnO}_4$ has highest amount of solute.

Both Statement I and II are correct.

Given the following information:

The vapor pressure of benzene at 300 K is $ p_{\text{Benzene}}^{\circ} = 9.7 \, \text{kPa} $.

The vapor pressure of toluene at 300 K is $ p_{\text{Toluene}}^{\circ} = 3.63 \, \text{kPa} $.

The mole fraction of toluene in the solution is $ x_{\text{Toluene}} = 0.4 $.

We need to determine the composition of the vapor in equilibrium with the solution, assuming it is an ideal solution.

Calculations:

Identify the mole fraction of benzene:

$ x_{\text{Benzene}} = 1 - x_{\text{Toluene}} = 1 - 0.4 = 0.6 $

Calculate the partial pressures using Raoult's Law:

Partial pressure of benzene:

$ p_{\text{Benzene}} = p_{\text{Benzene}}^{\circ} \times x_{\text{Benzene}} = 9.7 \, \text{kPa} \times 0.6 = 5.82 \, \text{kPa} $

Partial pressure of toluene:

$ p_{\text{Toluene}} = p_{\text{Toluene}}^{\circ} \times x_{\text{Toluene}} = 3.63 \, \text{kPa} \times 0.4 = 1.452 \, \text{kPa} $

Calculate the total pressure of the solution:

$ p_{\text{Total}} = p_{\text{Benzene}} + p_{\text{Toluene}} = 5.82 \, \text{kPa} + 1.452 \, \text{kPa} = 7.272 \, \text{kPa} $

Determine the mole fraction of toluene in the vapor phase using Dalton's Law:

$ Y_{\text{Toluene}} = \frac{p_{\text{Toluene}}}{p_{\text{Total}}} = \frac{1.452}{7.272} \approx 0.20 $

Conclusion:

The mole fraction of toluene in the vapor phase is approximately $ 0.20 $. Therefore, the composition of the vapor in equilibrium with the solution is $\approx 0.20$ for toluene.

Given:

Moles of solute, $ n_1 = 0.1 $ mol

Moles of solvent, $ n_2 = 0.9 $ mol

Vapor pressure of pure solvent, $ p^0 = 0.9 $ bar

The relative lowering of vapor pressure is calculated using the formula:

$ \frac{p^0 - p}{p^0} = \frac{n_1}{n_1 + n_2} $

Where:

$ p^0 $ is the vapor pressure of the pure solvent

$ n_1 $ is the number of moles of solute

$ n_2 $ is the number of moles of solvent

Substitute the given values into the equation:

$ \frac{0.9 - p}{0.9} = \frac{0.1}{0.1 + 0.9} $

Simplifying the right side:

$ \frac{0.1}{1.0} = 0.1 $

Now, plug this value back into the equation:

$ \frac{0.9 - p}{0.9} = 0.1 $

This implies:

$ 0.9 - p = 0.09 $

Solving for $ p $:

$ p = 0.9 - 0.09 = 0.81 \, \text{bar} $

Thus, the vapor pressure of the solution is $ 0.81 \, \text{bar} $.

Given,

Pressure (vapour) of pure solvent $ \left(\mathrm{H}_{2} \mathrm{O}\right) $ $ p^{\circ}=3.78 \mathrm{kPa} $.

Vapour pressure after solute mixing $ p=3.768 \mathrm{kPa} $.

Relative lowering of vapour pressure

$ =\frac{p^{\circ}-p}{p} $

$ =\frac{3.78-3.768}{3.78}=3.17 \times 10^{-3} $

Now, relative lowing of vapour pressure is given by

$ \frac{p^{\circ}-p}{p^{\circ}}=\frac{W_{B}-M_{A}}{M_{B} \times W_{A}} $

where,

$ W_{A}= $ mass of solvent $ \left(\mathrm{H}_{2} \mathrm{O}\right)(1 \mathrm{~kg}) $

$ M_{A}= $ Molar mass of solvent

$ W_{B}= $ Mass of solute ( 0.06 kg )

$ M_{B}= $ Molar mass of solvent

Substitute the values in Eq. (ii),

$ \begin{array}{c} 3.17 \times 10^{-3}=\frac{60 \mathrm{~g} \times 18 \mathrm{~g} / \mathrm{mol}}{M_{B} \times 1000 \mathrm{~g}} \\\\ M_{B}=340 \mathrm{~g} \mathrm{~mol}^{-1} \end{array} $

Given, percentage of nitrogen

$ \left(\mathrm{N}_2\right)=79 \% $

Percentage of oxygen $\left(\mathrm{O}_2\right)=21 \%$

Pressure $=1$ atm

Partial pressure of $\mathrm{N}_2$ and $\mathrm{O}_2$ are

$ \begin{aligned} p_{\mathrm{N}_2} & =p \times \mathrm{N}_2 \\\\ \Rightarrow \quad \frac{1 \times 79}{100} & =0.79 \mathrm{~atm} \quad \ldots \text { (i) } \\\\ p_{\mathrm{O}_2} & =p \times \mathrm{O}_2 \\\\ \Rightarrow \quad \frac{1 \times 21}{100} & =0.21 \mathrm{~atm} \quad \ldots \text { (ii) } \end{aligned} $

Applying Henry's law for $\mathrm{N}_2$ and $\mathrm{O}_2$,

$ \begin{aligned} p_{\mathrm{N}_2} & =K_{\mathrm{N}_2} x_{\mathrm{N}_2} \\\\ \Rightarrow \quad x_{\mathrm{N}_2} & =\frac{p_{\mathrm{N}_2}}{K_{\mathrm{N}_2}}=\frac{0.79}{8.57 \times 10^4} \\\\ x_{\mathrm{N}_2} & =9.25 \times 10^{-6} \quad \ldots \text { (iii) } \\\\ p_{\mathrm{O}_2} & =K_{\mathrm{O}_2} x_{\mathrm{O}_2} \\\\ \Rightarrow \quad x_{\mathrm{O}_2} & =\frac{p_{\mathrm{O}_2}}{K_{\mathrm{O}_2}} \\\\ & =\frac{0.21}{8.56 \times 10^4} \\\\ x_{\mathrm{O}_2} & =4.60 \times 10^{-6} \quad \ldots \text { (iv) } \end{aligned} $

Ratio of mole fraction of $\mathrm{N}_2$ and $\mathrm{O}_2$,

$ \frac{x_{\mathrm{N}_2}}{x_{\mathrm{O}_2}}=\frac{9.25 \times 10^{-6}}{4.60 \times 10^{-6}}=2.1 $

The relation between $K_f$ and $K_b$ is

$ K_f=K_b \times \frac{\Delta T_f}{\Delta T_b}\quad \ldots \text { (i) } $

$ \begin{aligned} & \Delta T_b=\text { elevation in boiling point } \\ & =373.202-373.15=0.052 \mathrm{~K} \end{aligned} $

$K_f$ and $K_b$ are given as,

$ \begin{aligned} & K_b=0.052 \mathrm{~K} \mathrm{~kg} / \mathrm{mol} \\ & K_f=1.86 \mathrm{~K} \mathrm{~kg} / \mathrm{mol} \end{aligned} $

Substitute all value in Eq. (i)

$ \begin{aligned} & 1.86=0.052 \times \frac{\Delta T_f}{0.52} \\ & \Delta T_f=\frac{1.86}{10} \text { or } 0.186\quad \ldots \text { (ii) } \end{aligned} $

$ \begin{aligned} & \Delta T_f=\text { depression in freezing point } \\ & \Delta T_f=0.186=273.15 \mathrm{~K}-x \\ & x=272.964 \mathrm{~K} \end{aligned} $

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 .

$\because$ 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.