Solutions

For an ideal solution, by Raoult’s law:

$P_{\text{total}}=x_A P_A^0 + x_B P_B^0$

First solution (3 mol $A$, 1 mol $B$)

Total moles $=3+1=4$

$x_A=\frac{3}{4},\quad x_B=\frac{1}{4}$

Given $P_{\text{total}}=500$ mm Hg:

$500=\frac{3}{4}P_A^0+\frac{1}{4}P_B^0$

Multiply by 4:

$2000=3P_A^0+P_B^0 \quad ...(1)$

After adding 1 mol $A$ (4 mol $A$, 1 mol $B$)

Total moles $=4+1=5$

$x_A=\frac{4}{5},\quad x_B=\frac{1}{5}$

New vapour pressure $=500+20=520$ mm Hg:

$520=\frac{4}{5}P_A^0+\frac{1}{5}P_B^0$

Multiply by 5:

$2600=4P_A^0+P_B^0 \quad ...(2)$

Solve

Subtract (1) from (2):

$2600-2000=(4P_A^0+P_B^0)-(3P_A^0+P_B^0)$

$600=P_A^0$

Put in (1):

$2000=3(600)+P_B^0=1800+P_B^0$

$P_B^0=200 \text{ mm Hg}$

Answer: $200$

Osmotic pressure ($\pi$) = $12 \text{ atm}$

Temperature ($T$) = $300 \text{ K}$

Gas constant ($R$) = $0.08 \text{ L atm K}^{-1} \text{ mol}^{-1}$

Sodium chloride (NaCl) is an electrolyte that breaks into ions in an aqueous solution:

$ \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- $

Since it is given to assume complete dissociation, one unit of NaCl produces 2 ions.

Therefore, the van 't Hoff factor, $i = 2$.

According to the formula for osmotic pressure:

$ \pi = iCRT $

Substitute the given values into the equation:

$ 12 = 2 \times C \times 0.08 \times 300 $

$ 12 = 2 \times 24 \times C $

$ 12 = 48 \times C $

$ C = \frac{12}{48} $

$ C = 0.25 \text{ mol L}^{-1} $

Molar mass of NaCl = Molar mass of Na + Molar mass of Cl

$ \text{Molar mass of NaCl} = 23 + 35.5 = 58.5 \text{ g mol}^{-1} $

To convert the concentration from moles per liter to grams per liter (strength), multiply the molar concentration ($C$) by the molar mass:

$ \text{Strength} (\text{g L}^{-1}) = \text{Molar concentration} \times \text{Molar mass} $

$ \text{Strength} = 0.25 \text{ mol L}^{-1} \times 58.5 \text{ g mol}^{-1} $

$ \text{Strength} = 14.625 \text{ g L}^{-1} $

Rounding off to the nearest integer, we get $15$.

The strength of the sodium chloride solution is 15 $\text{g L}^{-1}$.

$ \begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m} \\ & 0.5=\mathrm{i} \times \mathrm{m} \times 5 \\ & \mathrm{i} \times \mathrm{m}=\frac{0.5}{5}=0.1 \\ & \mathrm{i} \times \mathrm{a}=\frac{15}{1000} \end{aligned} $

(where $\mathrm{a}=$ moles of solute)

Now,

$\begin{aligned} & \frac{P_o-P_S}{P^o}=i X_{\text {solute }}=i \times \frac{a}{a+\frac{150}{300}} \\ & =i \times \frac{a}{1 / 2}=\frac{15 / 1000}{1 / 2}=\frac{30}{1000}=3 \times 10^{-2}=3\end{aligned}$

Sea water is 6 Molar in NaCl , So 1000 ml of sea water contains 6 mol of NaCl .

$\begin{aligned} & \text { mass of solution }=\text { Volume } \times \text { density } \\ & =1000 \times 2 \\ & \text { mass of solution }=2000 \mathrm{~g} \\ & \mathrm{ppm}=\frac{\text { mass of } \mathrm{O}_2}{2000} \times 10^6 \\ & \text { mass of } \mathrm{O}_2=5.8 \times 2 \times 10^{-3} \\ & \quad=1.16 \times 10^{-2} \mathrm{~g} \\ & \text { molality for } \mathrm{O}_2=\frac{1.16 \times 10^{-2} / 32}{(2000-6 \times 58.5)} \times 1000 \end{aligned}$

$\begin{aligned} &\begin{aligned} & =\frac{1.16 \times 10}{32 \times 1649} \\ & =0.000219 \\ & =2.19 \times 10^{-4} \end{aligned}\\ &\text { Correct answer } \Rightarrow 2 \end{aligned}$

For AB

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=2.7 \mathrm{~K} \\ & 2.7=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=\frac{27}{5} \end{aligned}$

Let molar mass of $A B=x$.

So $\frac{1 / x}{15} \times 1000=\frac{27}{5}$

$x=12.34$

For $\mathrm{AB}_2$

$\begin{aligned} & \Delta \mathrm{T}_{\mathrm{b}}=1.5 \mathrm{~K} \\ & 1.5=1 \times 0.5 \times \mathrm{m} \\ & \mathrm{~m}=3 \end{aligned}$

Let molar mass of $\mathrm{AB}_2=\mathrm{y}$

So $\frac{1 / \mathrm{y}}{15} \times 1000=3$

$\begin{aligned} & y=\frac{1000}{45} \\ & y=22.22 \end{aligned}$

Now let a and b be atomic masses of A and B respectively, then

$\begin{aligned} & \mathrm{A}+\mathrm{b}=12.34 \quad\text{...... (i)}\\ & \mathrm{~A}+2 \mathrm{~b}=22.22 \quad\text{...... (ii)}\\ & \mathrm{~B}=22.22-12.34=9.88 \end{aligned}$

Now $\mathrm{a}=12.34-9.88=2.46$

$=24.6 \times 10^{-1}=25 \times 10^{-1}$

Percent ionisation of $A_2B=30\%$

The dissociation of $A_2B$ in aqueous solution can be represented as

$A_2B \rightarrow 2A^+ + B^{2-}$

1 mole of $A_2B$ produces 2 moles of $A^+$ and 1 mole of $B^{2-}$ ions.

The total number of moles of ions produced (n) from the dissociation is

$n=2+1=3$

The degree of dissociation given that $A_2B$ is 30% ionised, the degree of dissociation $(\lambda)$ or degree of ionisation

$\lambda=\frac{30}{100}=0.3$

Van't Hoff factor ($i$) can be calculated using the formula,

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

Substitute the values of $n$ and $\lambda$ as,

$i = 1 + (3 - 1) \times 0.3$

$ = 1 + 2 \times 0.3$

$ = 1 + 0.6$

$ = 1.6$

$ = 16 \times {10^{ - 1}}$

To find the mole fraction of methyl benzene in the vapor phase when it is in equilibrium with an equimolar (equal mole) mixture of benzene and methyl benzene at $27^{\circ}C$, we can apply Raoult's law for an ideal solution. Given the vapor pressures of pure benzene and methyl benzene are 80 Torr and 24 Torr, respectively.

Raoult's law states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution ($P_i = P_i^\circ x_i$), where $P_i$ is the partial vapor pressure of the component $i$, $P_i^\circ$ is the vapor pressure of the pure component $i$, and $x_i$ is the mole fraction of the component $i$ in the solution.

For an equimolar mixture of benzene and methyl benzene, the mole fractions of benzene ($x_b$) and methyl benzene ($x_{mb}$) in the liquid phase are both $0.5$, because the mixture is equimolar.

The total pressure of the mixture ($P_{total}$) can be calculated using Raoult's law for each component and summing up the individual pressures:

$P_{total} = P_{benzene} + P_{methyl\ benzene}$

$P_{total} = (P_b^\circ x_b) + (P_{mb}^\circ x_{mb})$

Substituting the given values:

$P_{total} = (80 \times 0.5) + (24 \times 0.5)$

$P_{total} = 40 + 12 = 52 \, \text{Torr}$

The mole fraction of methyl benzene in the vapor phase ($y_{mb}$) can be calculated using Dalton's law, which states that the partial pressure of a component in a mixture is equal to the mole fraction of that component in the vapor phase times the total pressure of the mixture.

The partial pressure of methyl benzene can be obtained from Raoult's law as we did earlier:

$P_{methyl\ benzene} = P_{mb}^\circ x_{mb} = 24 \times 0.5 = 12 \, \text{Torr}$

Using Dalton's law:

$y_{mb} = \frac{P_{methyl\ benzene}}{P_{total}}$

Substituting the values:

$y_{mb} = \frac{12}{52}$

$y_{mb} = \frac{6}{26}$

$y_{mb} = \frac{3}{13}$

To express $y_{mb}$ as a percentage times $10^{-2}$:

$y_{mb} = \left(\frac{3}{13}\right) \times 100 \times 10^{-2}$

$y_{mb} \approx 23.08 \times 10^{-2}$

So, the mole fraction of methyl benzene in the vapor phase, in equilibrium with an equimolar mixture of those two liquids at $27^{\circ}C$, is approximately $23 \times 10^{-2}$ (rounded to the nearest integer).

To find the degree of ionization $(\alpha)$, we need to use the boiling point elevation formula and the van't Hoff factor. The formulas we will use are:

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

where:

$\Delta T_b$ is the boiling point elevation,

$i$ is the van't Hoff factor,

$K_b$ is the molal boiling point elevation constant, and

$m$ is the molality of the solution.

Given data:

• $\Delta T_b = 100.52^{\circ} \mathrm{C} - 100^{\circ} \mathrm{C} = 0.52^{\circ} \mathrm{C}$
• Mass of $\mathrm{AB}_2$ = 10 g
• Mass of water = 100 g = 0.1 kg
• Molar mass of $\mathrm{AB}_2$ = 200 g/mol
• $K_b$ = 0.52 K kg mol^-1

First, we calculate the molality (m):

$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}$

$= \frac{\frac{10}{200}}{0.1} \:\mathrm{mol\ kg}^{-1}$

$= 0.5 \:\mathrm{mol\ kg}^{-1}$

Next, using the formula for boiling point elevation:

$0.52 = i \cdot 0.52 \cdot 0.5$

$i = \frac{0.52}{0.52 \cdot 0.5}$

$i = \frac{1}{0.5}$

$i = 2$

Now, the van't Hoff factor $i$ is related to the degree of ionization $(\alpha)$. For the electrolyte $\mathrm{AB}_2$, which ionizes as $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$, the van't Hoff factor $i$ can be expressed as:

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

where

• $n$ is the number of ions produced (which is 3 for $\mathrm{AB}_2 \rightarrow \mathrm{A}^{2+}+2\mathrm{B}^{-}$)
• $\alpha$ is the degree of ionization

Substituting $i = 2$ into the equation:

$2 = 1 + 2 \alpha$

$2 - 1 = 2 \alpha$

$\alpha = \frac{1}{2}$

$\alpha = 0.5$

The degree of ionization $(\alpha)$ is therefore

$0.5 \times 10^{-1} = 5 \times 10^{-2}$.

Hence, the degree of ionization of the electrolyte $(\alpha)$ is approximately 5 $\times 10^{-1}$ (the nearest integer is 5).

$\begin{aligned} & \Delta T_f=2.5^{\circ} \mathrm{C} \\ & \Delta T_f=i \times k_f \times m \\ & 2.5=1 \times 1.86 \times \frac{n_B}{0.1} \\ & n_B=\frac{2.5 \times 0.1}{1.86}=0.1344 \mathrm{~mol} \\ & \text { mass of methanol }=0.1344 \times 32=4.3 \mathrm{~g} \\ & d=\frac{m}{v} \\ & v=\frac{m}{d} \\ & v=\frac{4.3}{0.792} \mathrm{~mL} \\ & =5.43 \mathrm{~mL} \\ & =543 \times 10^{-2} \mathrm{~mL} \\ & x=543 \end{aligned}$

To solve the problem, we'll calculate the freezing point depression of the acetic acid solution in water.

1. Calculating Moles of Acetic Acid:

• Given volume of acetic acid: $ 5 \, \text{mL} $


• Density of acetic acid: $ 1.2 \, \text{g/mL} $


• Molar mass of acetic acid: $ 60 \, \text{g/mol} $

$ \text{Mass of acetic acid} = \text{Volume} \times \text{Density} = 5 \, \text{mL} \times 1.2 \, \text{g/mL} = 6 \, \text{g} $

$ \text{Moles of acetic acid} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} $

1. Calculating Molality:

• Volume of water: $ 1 \, \text{L} $


• Density of water: $ 1 \, \text{g/cm}^3 \Rightarrow 1 \, \text{kg/L} $


• Hence, mass of water = $ 1 \, \text{kg} $

$ \text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent (kg)}} = \frac{0.1 \, \text{mol}}{1 \, \text{kg}} = 0.1 \, \text{mol/kg} $

1. Degree of Dissociation ($\alpha$):

• Dissociation constant ($K_a$) of acetic acid: $6.25 \times 10^{-5}$

$ \alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{6.25 \times 10^{-5}}{0.1}} = \sqrt{6.25 \times 10^{-4}} = 25 \times 10^{-3} = 0.025 $

1. Van't Hoff factor (i):

• Acetic acid partially dissociates into 2 ions (CH$_3$COO$^-$ and H$^+$)

$ i = 1 + (\text{number of ions produced} - 1) \times \alpha = 1 + (2-1) \times 0.025 = 1 + 0.025 = 1.025 $

1. Freezing Point Depression ($\Delta T_f$):

• Freezing point depression constant ($K_f$) for water: $1.86 \,\text{K kg/mol}$

$ \Delta T_f = i \times K_f \times \text{Molality} = 1.025 \times 1.86 \times 0.1 = 0.19065 \, \text{K} $

1. Final Freezing Point:

• Pure water freezes at $0 \, ^\circ \text{C}$

$ \text{Freezing point of solution} = 0 \, ^\circ \text{C} - 0.19065 \, ^\circ \text{C} = -0.19065 \, ^\circ \text{C} $

Here, $-x \times 10^{-2} \, ^\circ \text{C}$ is given. Therefore, $x = 19$ (nearest integer).

So, the final answer is:

$ x = 19 $

An artificial cell contains a 0.2 M glucose solution within a semipermeable membrane. When this cell is placed in a 0.05 M NaCl solution at 300 K, we need to determine the osmotic pressure developed, expressed as × 10⁻¹ bar.

Given:

Ideal Gas Constant, R = 0.083 L bar mol⁻¹ K⁻¹

Assume complete dissociation of NaCl.

Explanation

The osmotic pressure ($\pi$) is calculated using the formula:

$ \pi = \left( i_1 C_1 - i_2 C_2 \right) R T $

where:

$i_1$ and $C_1$ are the van 't Hoff factor and concentration for glucose.

$i_2$ and $C_2$ are the van 't Hoff factor and concentration for NaCl.

$R$ is the ideal gas constant.

$T$ is the temperature in Kelvin.

For glucose:

$i_1 = 1$ (glucose does not dissociate)

$C_1 = 0.2$ M

For NaCl:

$i_2 = 2$ (NaCl dissociates into two ions, Na⁺ and Cl⁻)

$C_2 = 0.05$ M

Plugging in the values:

$ \pi = \left(1 \times 0.2 - 2 \times 0.05\right) \times 0.083 \times 300 $

Calculating:

$ \pi = (0.2 - 0.1) \times 0.083 \times 300 $

$ \pi = 0.1 \times 0.083 \times 300 $

$ \pi = 2.5 \text{ bar} $

Hence, the osmotic pressure developed is $25 \times 10^{-1}$ bar.

$\begin{aligned} & \text { Molality of acetic acid }=\frac{2700}{60} \times \frac{1}{2.7} \mathrm{~mol} / \mathrm{kg} \\ &=16.667 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \times 16.667 \\ &=1.86 \times 16.667 \\ &=31 \mathrm{~K} \end{aligned}$

• Molal boiling point elevation constant of water ($K_b$) = 0.52 K.kg.mol^-1
• 1 atm = 760 mm Hg
• Molar mass of water = 18 g.mol^-1

First, we calculate the molality (m) of the solution using the boiling point elevation formula:

$ \Delta T_b = K_b \times m $

From the problem, we know:

$ 2 = 0.52 \times m $

Solving for m:

$ m = \frac{2}{0.52} \approx 3.846 \text{ mol/kg} $

Next, considering the solution is highly diluted, we use the formula for relative lowering of vapor pressure:

$ \frac{\Delta P}{P^0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} $

Given that molality (m) is defined as the number of moles of solute per kilogram of solvent:

$ \frac{\Delta P}{P^0} = \frac{m}{1000} \times M_{solvent} $

Substitute the known values:

$ \Delta P = P^0 \times \frac{m}{1000} \times M_{solvent} $

$ = 760 \times \frac{3.846}{1000} \times 18 $

$ = 52.615 \text{ mm Hg} $

Therefore, the vapor pressure of the solution:

$ P_{solution} = P^0 - \Delta P $

$ = 760 - 52.615 $

$ \approx 707.385 \text{ mm Hg} $

Rounding to the nearest integer, the vapor pressure of the aqueous solution is 707 mm Hg.

To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to $-24^{\circ} \mathrm{C}$, we can use the freezing point depression equation, which is a colligative property given by:

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

where:

• $\Delta T_f$ is the depression in the freezing point.
• $i$ is the van't Hoff factor, which is the number of particles the solute splits into in solution. For ethylene glycol, which does not dissociate in solution, $i = 1$.
• $K_f$ is the cryoscopic constant (freezing point depression constant) of water, which is given as $1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}$.
• $m$ is the molality of the solution.

However, molality is defined as the number of moles of solute per kilogram of solvent, so we first need to find the molality that corresponds to the desired freezing point depression:

$ \Delta T_f = -24^{\circ} \mathrm{C} - (0^{\circ} \mathrm{C}) = -24^{\circ} \mathrm{C} $

Now, we can solve for $m$:

$ m = \frac{\Delta T_f}{i \cdot K_f} $

Substituting the given values:

$ m = \frac{-24^{\circ} \mathrm{C}}{1 \cdot 1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}} $

Note that $\mathrm{C}$ and $\mathrm{K}$ are interchangeable when calculating changes in temperature. Now, we can compute the molality:

$ m = \frac{-24}{1.86} $

Next, we calculate the molality of the ethylene glycol:

$ m \approx -12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} $

Keep in mind that molality is always positive, but the negative sign indicates the direction of the temperature change. Since we're calculating the amount needed, we can consider it as 12.903 moles per kilogram of water.

Now we calculate the moles of ethylene glycol required for $18.6 \mathrm{~kg}$ of water:

$ \text{moles of ethylene glycol} = m \cdot \text{mass of water (in kg)} $ $ \text{moles of ethylene glycol} = 12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} \cdot 18.6 \mathrm{~kg} $ $ \text{moles of ethylene glycol} \approx 240.0158 \mathrm{~mol} $

Next, we find the mass of ethylene glycol needed using its molar mass:

$ \text{mass of ethylene glycol} = \text{moles of ethylene glycol} \cdot \text{molar mass of ethylene glycol} $

Given the molar mass of ethylene glycol is $62 \mathrm{~g} ~\mathrm{mol}^{-1}$, we then convert it to $\mathrm{kg} ~\mathrm{mol}^{-1}$ by dividing by 1000:

$ \text{molar mass of ethylene glycol} = 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} $

Now we can calculate the mass:

$ \text{mass of ethylene glycol} = 240.0158 \mathrm{~mol} \cdot 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} $ $ \text{mass of ethylene glycol} \approx 14.88098 \mathrm{~kg} $

Therefore, approximately $14.881 \mathrm{~kg}$ of ethylene glycol needs to be added to $18.6 \mathrm{~kg}$ of water to lower the freezing point to $-24^{\circ} \mathrm{C}$.

To calculate the osmotic pressure of the solution at the new temperature, we can use the formula:

$ \pi = \text{CRT} $

Since the concentration $ C $ and the gas constant $ R $ remain constant, the ratio of the osmotic pressures at two different temperatures is given by:

$ \frac{\pi_1}{\pi_2} = \frac{T_1}{T_2} $

Thus, we can express the osmotic pressure at the second temperature as:

$ \pi_2 = \frac{\pi_1 \cdot T_2}{T_1} $

Substituting the given values:

$ \pi_2 = \frac{7 \times 10^5 \times 283}{273} $

After calculation, this simplifies to:

$ \pi_2 = 72.56 \times 10^4 \, \text{Nm}^{-2} $

The given solution is a $30 \%$ (w/v) aqueous solution of glucose. This means that $30\ \mathrm{g}$ of glucose is dissolved in $100\ \mathrm{mL}$ of the solution. Since the density of the solution is $1.2\ \mathrm{g/mL}$, the weight of $100\ \mathrm{mL}$ of the solution is:

$\mathrm{Wt.~of~solution} = 100\ \mathrm{mL} \times 1.2\ \mathrm{g/mL} = 120\ \mathrm{g}$

Since the solution is $30 \%$ (w/v), the weight of glucose in $100\ \mathrm{mL}$ of the solution is $30\ \mathrm{g}$. Therefore, the weight of water in $100\ \mathrm{mL}$ of the solution is:

$\mathrm{Wt.~of~water} = 120\ \mathrm{g} - 30\ \mathrm{g} = 90\ \mathrm{g}$

Now, we can use Raoult's law to calculate the vapour pressure of the solution. Raoult's law states that the partial vapour pressure of a solvent in a solution is proportional to its mole fraction in the solution. For a dilute solution, the mole fraction of the solvent can be approximated as the ratio of the number of moles of solvent to the total number of moles in the solution.

Let us assume that we have 1 mole of the solution. The number of moles of glucose in the solution is:

$n_{\mathrm{glucose}} = \frac{30\ \mathrm{g}}{180\ \mathrm{g/mol}} = 0.167\ \mathrm{mol}$

The number of moles of water in the solution is:

$n_{\mathrm{water}} = \frac{90\ \mathrm{g}}{18\ \mathrm{g/mol}} = 5\ \mathrm{mol}$

The total number of moles in the solution is:

$n_{\mathrm{total}} = n_{\mathrm{glucose}} + n_{\mathrm{water}} = 5.167\ \mathrm{mol}$

Let $P$ be the vapour pressure of the solution. According to Raoult's law, we have:

$\frac{P_0 - P}{P} = \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}} = \frac{0.167}{5} = 0.0334$

where $P_0$ is the vapour pressure of pure water, which is given as $24\ \mathrm{mmHg}$.

Simplifying the above equation, we get:

$P = \frac{P_0}{1 + \frac{n_{\mathrm{glucose}}}{n_{\mathrm{water}}}} = \frac{24\ \mathrm{mmHg}}{1 + \frac{0.167}{5}} = 23.22\ \mathrm{mmHg}$

Therefore, the vapour pressure of the $30\%$ (w/v) aqueous solution of glucose at $25^\circ\mathrm{C}$ is $\boxed{23\ \mathrm{mmHg}}$ (rounded off to the nearest integer).

Amount of solvent = 100 - (29.25 + 19) = 51.75 g

Now, we can calculate the boiling point elevation using the given formula:

ΔTb = [(2 × 29.25 × 1000) / (58.5 × 51.75) + (3 × 19 × 1000) / (95 × 51.75)] × 0.52

ΔTb = 16.075

The boiling point of the sea water is the normal boiling point of water plus the change in boiling point:

Boiling point of sea water = 100 °C + 16.075 °C = 116.075 °C

Rounding to the nearest integer, the normal boiling point of the sea water is approximately 116 °C.

To solve this problem, we will first calculate the osmotic pressure of the 0.05 M glucose solution and then use that information to determine the molecular mass of compound A.

1. Osmotic pressure equation:

$\Pi = iMRT$

where $\Pi$ is the osmotic pressure, $i$ is the van't Hoff factor (which is 1 for non-electrolytes), $M$ is the molarity, $R$ is the ideal gas constant (0.0821 L atm/mol K), and $T$ is the temperature in Kelvin.

Since both solutions have the same osmotic pressure at the same temperature, we can set their osmotic pressures equal to each other:

$\Pi_{A} = \Pi_{glucose}$

2. Calculate the osmotic pressure of the 0.05 M glucose solution:

Glucose is a non-electrolyte, so its van't Hoff factor is 1. We don't know the temperature, but since both solutions are at the same temperature, it will cancel out in our calculations.

$\Pi_{glucose} = (1)(0.05 ~\mathrm{M})(R)(T)$

3. Calculate the molarity of compound A:

Since we know that 12 g of compound A is dissolved in 1000 mL of water, we can find the molarity once we know the molecular mass.

Let $x$ be the molecular mass of compound A. Then, the molarity of compound A is:

$M_{A} = \frac{12 ~\mathrm{g}}{x ~\mathrm{g/mol}} \times \frac{1}{1 ~\mathrm{L}} = \frac{12}{x} ~\mathrm{M}$

4. Set the osmotic pressures equal to each other:

$\Pi_{A} = \Pi_{glucose}$

$(1)(\frac{12}{x} ~\mathrm{M})(R)(T) = (1)(0.05 ~\mathrm{M})(R)(T)$

The van't Hoff factors, ideal gas constant, and temperature cancel out:

$\frac{12}{x} = 0.05$

5. Solve for the molecular mass (x) of compound A:

$x = \frac{12}{0.05}$

$x = 240 ~\mathrm{g/mol}$

The molecular mass of compound A is 240 g/mol.

$ \begin{array}{ccc} \mathrm{MgCl}_2 & \rightleftharpoons ~\mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-} \\ 1 & 0 & 0 \\ 1-0.8 & 0.8 & 1.6 \end{array} $

Hence overall molality will be equal to $=2.6$

$ \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}+2.6} $

For dil solution

$ \begin{aligned} & \frac{p^{\circ}-p}{p^{\circ}}=\frac{2.6}{\frac{1000}{18}} \\\\ & p=47.66 \simeq 48 \mathrm{~mm} \mathrm{Hg} \end{aligned} $

Given that isotonic solutions have the same osmotic pressure, we equate the osmotic pressures of the K₂SO₄ and glucose solutions :

$i \times 0.004 \, \text{M} = 0.01 \, \text{M}$

Here, $i$ is the van't Hoff factor, which accounts for the number of ions the compound dissociates into in the solution. Solving this equation for $i$, we get $i = 2.5$.

For the compound K₂SO₄, which dissociates into 3 ions (2K⁺ and 1 SO₄^2-), the formula for $i$ is given by

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

where $n$ is the number of ions the solute dissociates into and $\alpha$ is the degree of dissociation.

Substituting $n=3$ and $i=2.5$ into this formula and solving for $\alpha$ gives

$\alpha = \frac{i-1}{n-1} = \frac{2.5-1}{3-1} = 0.75$

To convert this into a percentage, we multiply by 100, yielding a percentage dissociation of 75%.

The concept we are utilizing to solve this problem is osmotic pressure. Osmotic pressure is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. For dilute solutions, osmotic pressure behaves similarly to an ideal gas, hence the formula we use is similar to the ideal gas law:

$\Pi = \frac{n}{V}RT$

where:

• $\Pi$ is the osmotic pressure,
• $n$ is the number of moles of solute,
• $V$ is the volume of the solution,
• $R$ is the ideal gas constant, and
• $T$ is the temperature in Kelvin.

First, we convert all given quantities to the appropriate units.

• $\Pi = 1.29$ mbar is equivalent to $1.29 \times 10^{-3}$ bar (since 1 bar = 1000 mbar),
• $V = 300$ cm³ is equivalent to $0.3$ L (since 1 L = 1000 cm³),
• $T = 300$ K,
• $R = 0.083$ L bar K$^{-1}$ mol$^{-1}$.

Instead of the number of moles, we are given the mass of the solute. However, we can replace the moles with mass using the relationship $n = \frac{m}{M}$, where $M$ is the molar mass, and $m$ is the mass of the solute.

So the equation becomes:

$\Pi = \frac{m}{MV}RT$

We are trying to solve for $M$, so rearrange the equation to isolate $M$:

$M = \frac{mRT}{\Pi V}$

Finally, substitute the given values into the equation:

$M = \frac{0.63 \times 0.083 \times 300}{1.29 \times 10^{-3} \times 0.3} \approx 40535 \, \text{g/mol}$

So the molar mass of the protein is approximately 40535 g/mol.

The degree of dissociation, often represented as $\alpha$, is the fraction of a mole of a substance that has dissociated into ions in solution. For a weak monobasic acid, this degree of dissociation can increase the number of particles in solution, which can in turn affect colligative properties such as the freezing point.

In this case, a weak monobasic acid, when it dissociates, produces two particles: one $H^{+}$ ion and one anion. So if the degree of dissociation is 0.3 ($\alpha = 0.3$), the average number of particles per molecule of the acid ($i$), also known as the van't Hoff factor, will be $1 + \alpha = 1 + 0.3 = 1.3$.

The decrease in freezing point ($\Delta T_{f}$) is given by the formula $\Delta T_{f} = i \cdot m \cdot K_{f}$, where $m$ is the molality of the solution and $K_{f}$ is the cryoscopic constant of the solvent. Therefore, the observed freezing point depression will be 1.3 times the theoretical freezing point depression for a non-dissociating solute (where $i = 1$).

So, the observed freezing point will be 30% higher than the theoretical freezing point, given the degree of dissociation of 0.3.

The boiling point elevation constant, also known as the ebullioscopic constant ($K_{b}$), can be determined using the formula:

$K_{b} = \frac{R \cdot T_{b}^{2}}{1000 \cdot \Delta H_{vap}}$

where:

- $R$ is the universal gas constant (8.31 J mol⁻¹ K⁻¹)

- $T_{b}$ is the boiling point of the solvent (in K)

- $\Delta H_{vap}$ is the enthalpy of vaporization (in J/mol)

According to the problem, the ratio of boiling points for X and Y is 2:1, and the ratio of their enthalpy of vaporization is 1:2.

Let's denote the boiling point of Y as $T_{b}(Y)$ and the boiling point of X as $T_{b}(X)$, and likewise for the enthalpy of vaporization $\Delta H_{vap}(Y)$ and $\Delta H_{vap}(X)$.

We then have:

$T_{b}(X) = 2T_{b}(Y)$ and $\Delta H_{vap}(X) = 0.5\Delta H_{vap}(Y)$

We can substitute these values into the equation for $K_{b}$:

$K_{b}(X) = R*(2T_{b}(Y))^2 / (1000 * 0.5\Delta H_{vap}(Y)) = 4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))$

and

$K_{b}(Y) = R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))$

Comparing these two equations:

$K_{b}(X) / K_{b}(Y) = [4*R*T_{b}(Y)^2 / (500 * \Delta H_{vap}(Y))] / [R*T_{b}(Y)^2 / (1000 * \Delta H_{vap}(Y))] $

This simplifies to:

$K_{b}(X) / K_{b}(Y) = 4 / 0.5 = 8$

So, the boiling point elevation constant of X is 8 times the boiling point elevation constant of Y. Therefore, m = 8.

At any temperature, the vapour pressure of the solution is lower than that of the pure solvent. Hence vapour pressure- temperature curve of solution lies below that of solvent.

The more volatile liquid evaporates fast as compared to the less volatile liquid at a low temperature because the volume increases with respect to temperature so it has a low boiling point.

We say that two solutions are isotonic (at the same temperature) if they have the same osmotic pressure, $\pi$. For dilute solutions at the same temperature $T$,

$ \pi = i\, M\, R\, T, $

where

$M$ = molarity of the solution,

$i$ = van 't Hoff factor (number of particles the solute dissociates into),

$R$ = universal gas constant,

$T$ = absolute temperature.

Since $R$ and $T$ are common for both solutions (same temperature), two solutions will be isotonic if

$ i_1 \, M_1 \;=\; i_2 \, M_2. $

Let us check each pair:

---

1. Pair A: $1\,\mathrm{M}$ NaCl and $2\,\mathrm{M}$ urea

NaCl dissociates into $ \mathrm{Na}^+ $ and $ \mathrm{Cl}^- $.

Hence $i(\mathrm{NaCl}) = 2$.

So $i \, M = 2 \times 1 = 2.$

Urea ($\mathrm{(NH_2)_2CO}$) does not dissociate in water.

Hence $i(\mathrm{urea}) = 1$.

So $i \, M = 1 \times 2 = 2.$

Both solutions have $i M = 2.$

$\therefore$ They are isotonic.

---

2. Pair B: $1\,\mathrm{M}$ $\mathrm{CaCl_2}$ and $1.5\,\mathrm{M}$ $\mathrm{KCl}$

Calcium chloride ($\mathrm{CaCl_2}$) dissociates into $ \mathrm{Ca}^{2+} $ and $ 2\,\mathrm{Cl}^- $.

Hence $i(\mathrm{CaCl_2}) = 3.$

So $i \, M = 3 \times 1 = 3.$

Potassium chloride ($\mathrm{KCl}$) dissociates into $ \mathrm{K}^+ $ and $ \mathrm{Cl}^- $.

Hence $i(\mathrm{KCl}) = 2.$

So $i \, M = 2 \times 1.5 = 3.$

Both solutions have $i M = 3.$

$\therefore$ They are isotonic.

---

3. Pair C: $1.5\,\mathrm{M}$ $\mathrm{AlCl_3}$ and $2\,\mathrm{M}$ $\mathrm{Na_2SO_4}$

Aluminum chloride ($\mathrm{AlCl_3}$) dissociates into $ \mathrm{Al}^{3+} $ and $ 3\,\mathrm{Cl}^- $.

Hence $i(\mathrm{AlCl_3}) = 4.$

So $i \, M = 4 \times 1.5 = 6.$

Sodium sulfate ($\mathrm{Na_2SO_4}$) dissociates into $ 2\,\mathrm{Na}^+ $ and $ \mathrm{SO_4}^{2-} $.

Hence $i(\mathrm{Na_2SO_4}) = 3.$

So $i \, M = 3 \times 2 = 6.$

Both solutions have $i M = 6.$

$\therefore$ They are isotonic.

---

4. Pair D: $2.5\,\mathrm{M}$ $\mathrm{KCl}$ and $1\,\mathrm{M}$ $\mathrm{Al_2(SO_4)_3}$

Potassium chloride ($\mathrm{KCl}$) has $i(\mathrm{KCl}) = 2.$

So $i \, M = 2 \times 2.5 = 5.$

Aluminum sulfate $\mathrm{Al_2(SO_4)_3}$ dissociates into

$ 2\,\mathrm{Al}^{3+} \;+\; 3\,\mathrm{SO_4}^{2-} \quad \Longrightarrow i(\mathrm{Al_2(SO_4)_3}) = 2 + 3 = 5. $

So $i \, M = 5 \times 1 = 5.$

Both solutions have $i M = 5.$

$\therefore$ They are isotonic.

---

Conclusion

All four given pairs $(A, B, C, D)$ satisfy $i_1 M_1 = i_2 M_2$. Therefore, all of them are isotonic pairs.

Hence, the number of isotonic pairs is:

$ \boxed{4}. $

Given:

• Vapor pressure reduction: $25\%$ ($0.75$ times the vapor pressure of pure water)


• Molar mass of water ($\text{H}_2\text{O}$): $18 \, \text{g/mol}$


• Mass of solvent (water): $1000 \, \text{g}$

Using Raoult's law:

$\frac{P^0 - P_s}{P_s} = \frac{n_{\text{solute}}}{n_{\text{solvent}}} = \frac{\frac{x}{M_{\text{urea}}}}{\frac{1000}{M_{\text{water}}}} = \frac{P^0 - 0.75P^0}{0.75P^0}$

Solving for (x):

$\frac{x}{60} = \frac{10000}{9}$ $x = 1111.11111 \approx 1111 \, \text{g}$

So, the mass of urea required to be dissolved in $1000 \, \text{g}$ of water is $1111 \, \text{g}$.

$\begin{aligned} & \mathrm{i}=1+(\mathrm{n}-1) \alpha \\\\ & \Rightarrow \mathrm{i}=1+0.2(2-1)=1.2 \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=1.2 \times 1.86 \times \frac{5 \times 1000}{60 \times 500} \\\\ & \Delta \mathrm{t}_{\mathrm{f}}=0.372 \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=372 \times 10^{-3}\end{aligned}$

$25 \times M=20 \times 1$

$ \begin{aligned} M & =\frac{20}{25}=\frac{4}{5}=0.8 \\\\ \Delta T_{f} & =(\mathrm{i})\left(\mathrm{K}_{\mathrm{f}}\right)(\mathrm{m}) \\\\ & =(2)(2)\left(\frac{4}{5}\right)=\frac{16}{5}=3.2 \end{aligned} $

$\pi = CRT$

$ \begin{aligned} 400 & =\frac{2.5}{\mathrm{m_w}} \times 4 \times\left(.083 \times 10^5\right) \times 300 \\\\ \mathrm{m_w} & =\frac{10 \times 0.083 \times 3}{4} \times 10^5 \\\\ & =62250 \end{aligned} $

$\mathrm{\Delta T_f=K_f~i~m}$

$ = 1.8 \times 2.67 \times {{{{38} \over {98}}} \over {0.062}} = 30$

$\therefore$ It freeze at $273-30=243$ $\mathrm{K}$

$\mathrm{\Delta T_b=K_b.m}$

(0.52) = (0.52) (m)

$\mathrm{m=1=\frac{2(1000)}{(mw)(20)}}$

mw = 100

$0.15=3 \times 0.5 \times \mathrm{M}$

$ \begin{aligned} & \mathrm{M}_{\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}}=0.1 \text { molar } \\\\ & \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NaCl} \longrightarrow \mathrm{PbCl}_{2}+2 \mathrm{NaNO}_{3} \\\\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \cdot \mathrm{m} \\\\ & 0.8=(0.4+3 s) 1.8 \\\\ & s=0.0148 \\\\ & \therefore \text { solubility product }=4 \mathrm{~s}^{3} \\\\ & =4 \times(0.0148)^{3} \\\\ & \approx 13 \times 10^{-6} \end{aligned} $

$\pi=i C R T$

The following pairs of solutions have same value of osmotic pressure

(A) $0.500 \mathrm{M} ~\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{aq}) \mathrm{i}=1$ and $0.25 \mathrm{M} ~\mathrm{KBr}(\mathrm{aq})$ $i=2$

$ \begin{aligned} & \pi_1=0.5 \times 1 \times R T =0.5 R T \\\\ \mathrm{KBr} & \rightleftharpoons \mathrm{K}^{+}+\mathrm{Br}^{-} \\\\ & \pi_2=0.25 \times 2 \times R T=0.5 R T \end{aligned} $

Thus, $\pi_1=\pi_2$

(B) $0.100 \mathrm{M} ~\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right](\mathrm{aq}) \mathrm{i}=5$ and $0.100 \mathrm{M}$ $\mathrm{FeSO}_{4}\left(\mathrm{NH}_{4}\right)_{2}(\mathrm{aq}) \mathrm{i}=5$

$ \begin{array}{ll} \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 \mathrm{~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \quad \pi_1=0.1 \times 5 \times R T=0.5 R T & \\\\ \mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \rightleftharpoons \mathrm{Fe}^{2+}+2 \mathrm{NH}_4^{+}+2 \mathrm{SO}_4^{2-} & (i=5) \\\\ \pi_2=0.1 \times 5 \times R T=0.5 R T & \end{array} $

Thus, $\pi_1=\pi_2$

C. 0.05 $\mathrm{M~K_4[Fe(CN)_6]~(aq)}$ and 0.25 $\mathrm{M~NaCl~(aq)}$

$ \begin{gathered} {K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightleftharpoons 4 {~K}^{+}+\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-} (i=5) \\\\ \pi_1=0.05 \times 5 \times R T=0.25 R T \\\\ {NaCl} \rightleftharpoons \mathrm{Na}^{+}+\mathrm{Cl}^{-} \quad(i=2) \\\\ \pi_2=0.25 \times 2 R T=0.5 R T \end{gathered} $

Thus, $\pi_1 \neq \pi_2$

(D) $0.15 \mathrm{M}~ \mathrm{NaCl}(\mathrm{aq}) \mathrm{i}=2$ and $0.10 \mathrm{M}~ \mathrm{BaCl}_{2}$ (aq) $i=3$

$ \pi_1=0.15 \times 2 \times R T=0.3 R T$

$ \begin{aligned} \mathrm{BaCl}_2 \rightleftharpoons \mathrm{Ba}^{2+}+2 \mathrm{Cl}^{-}(i=3) \\\\ \pi_2=0.1 \times 3 \times R T=0.3 R T \end{aligned} $

Thus, $\pi_1=\pi_2$

(E) $0.02 \mathrm{M}~ \mathrm{KCl} . \mathrm{MgCl}_{2} .6 \mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) \mathrm{i}=5$ and $0.05 \mathrm{M}$ $\mathrm{KCl}(\mathrm{aq}) \mathrm{i}=2$

$ \mathrm{KCl} \cdot \mathrm{MgCl}_2 \cdot 6 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{K}^{+}+3 \mathrm{Cl}^{-}+\mathrm{Mg}^{2+} \quad(i=5)$

$ \begin{aligned} & \pi_1=5 \times 0.02 R T=0.1 R T \\\\ & 0.05 \mathrm{M} \mathrm{KCl}, \mathrm{KCl} \rightleftharpoons \mathrm{K}^{+}+\mathrm{Cl}^{-}(i=2) \\\\ & \pi_2=0.05 \times 2 R T=0.1 R T \end{aligned} $

Thus, $\pi_1=\pi_2$

$\begin{aligned} & \pi=\mathrm{M}^{\prime} \mathrm{RT}=\left(\frac{\mathrm{W} / \mathrm{M}}{\mathrm{V}}\right) \mathrm{RT} \\\\ & \Rightarrow \ \pi=\left(\frac{\mathrm{W}}{\mathrm{V}}\right)\left(\frac{1}{\mathrm{M}}\right) \mathrm{RT}=\mathrm{C}\left(\frac{\mathrm{RT}}{\mathrm{M}}\right) \\\\ & \Rightarrow \ \frac{\pi}{\mathrm{C}}=\frac{\mathrm{RT}}{\mathrm{M}} =\mathrm{Constant}\end{aligned}$

If we assume graph between $\frac{\pi}{\mathrm{C}}$ and $\mathrm{C}$, then right graph should be like this in the question,



$ \therefore $ Question's graph is wrong.

Assuming $\pi $ vs C graph,

$\begin{aligned} & \pi=C R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times R T \\\\ & \Rightarrow \pi=\frac{\text { mole }}{\text { volume }} \times \frac{\mathrm{M}}{\mathrm{M}} \times \mathrm{RT} \\\\ & \Rightarrow \pi=\frac{\text { mass }}{\text { volume }} \times \frac{\mathrm{RT}}{\mathrm{M}} \\\\ & \Rightarrow \pi(\mathrm{atm})=\frac{R T}{M} \times C\left(\mathrm{g} \mathrm{lit}^{-1}\right)\end{aligned}$

$\begin{aligned} & \text {So, Slope }=\frac{\mathrm{RT}}{\mathrm{M}}=\frac{0.083 \times 300}{\mathrm{M}}=6 \times 10^{-4} \\\\ & \therefore \mathrm{M}=\frac{0.083 \times 300}{6 \times 10^{-4}}=\frac{830 \times 300}{6} \\\\ & =41,500 \mathrm{g} / \mathrm{mole}\end{aligned}$

Let $V . P$ of pure $A$ be $P_A^0$

Let $V . P$ of pure $B$ be $P_B^0$

When $\mathrm{X}_{\mathrm{A}}=0.7 \& \mathrm{X}_{\mathrm{B}}=0.3$

$ \begin{aligned} & \mathrm{P}_{\mathrm{s}}=350 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.7+\mathrm{P}_{\mathrm{B}}^0 \times 0.3=350 \end{aligned} $

When $\mathrm{X}_{\mathrm{A}}=0.2 \& \mathrm{X}_{\mathrm{B}}=0.8$

$ \begin{aligned} & P_{\mathrm{s}}=410 \\\\ & \Rightarrow \mathrm{P}_{\mathrm{A}}^0 \times 0.2+\mathrm{P}_{\mathrm{B}}^0 \times 0.8=410 \end{aligned} $

Solving (i) and (ii)

$ \begin{aligned} & \mathrm{P}_{\mathrm{A}}^0=314 \mathrm{~mm} \mathrm{Hg} \\\\ & \mathrm{P}_{\mathrm{B}}^0=434 \mathrm{~mm} \mathrm{Hg} \\\\ & \therefore \text{Answer} = 314 \end{aligned} $

Mass of solvent $=d x v=0.8 \times 62.5=50$ gram

$\Delta T_{f}=k_{f} \times m$

$ \begin{aligned} &0.9=2\left[\frac{1.8 \times 1000}{M_{\text {Solute }} \times 50}\right] \\ &M_{\text {Solute }}=\left(\frac{2 \times 1.8 \times 1000}{0.9 \times 50}\right)=80 \end{aligned} $

$\mathrm{P}=\mathrm{K}_{\mathrm{H}} \times \mathrm{X}$

$0.920 \mathrm{bar}=46.82 \times 10^{3}\, \mathrm{bar} \times \frac{\mathrm{mol} \,\mathrm{of} \,\mathrm{O}_{2}}{\mathrm{~mol} \text { of } \mathrm{H}_{2} \mathrm{O}}$

$0.920=46.82 \times 10^{3} \times \frac{\text { mol of O}_{2}}{1000 / 18}$

$0.920=46.82 \times n_{O_{2}}$

$\mathrm{P}=\frac{0.920}{46.82 \times 18}=n_{O_{2}}$

$\Rightarrow 1.09 \times 10^{-3} \,n_{O_{2}}$

$\Rightarrow \mathrm{m}\, \mathrm{mol}\, \mathrm{of}\, \mathrm{O}_{2}=1$

Given that $X_{A}=0.2, Y_{A}=0.5, P_{T}=0.8 \mathrm{~atm}$

We know that $P_{A}=Y_{A} \times P_{T}$

$ P_{A}=0.5 \times 0.8=0.4 $

Now $P_{A}=X_{A} \times P_{A}^{\circ} \Rightarrow P_{A}^{\circ}=\frac{0.4}{0.2}=2 \mathrm{~atm}$

M.wt. of Acetic acid $=60 \mathrm{~g}$

M.wt. of Ascorbic acid $=176 \mathrm{~g}$

$ \Delta T_{f}=K_{f} m $

$ \begin{aligned} \Delta T_{f} &=\frac{3.9 \times 10.2 \times 1000}{176 \times 150} \\\\ \Delta T_{f} &=1.506 \\\\ &=15.06 \times 10^{-1} \\\\ &=15 \end{aligned} $

$\because$ Diliute solution given:

$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}^0} \sim \frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $

$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $

${ }^{\mathrm{n}}$ solute $\sim \frac{{ }^{\mathrm{n}} \text { solvent }}{2}=\frac{100}{18 \times 2}=2.78 \mathrm{~mol}$

More accurate approach:

$ \frac{\mathrm{P}^0-\mathrm{P}_{\mathrm{S}}}{\mathrm{P}_{\mathrm{S}}}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^{\mathrm{n}} \text { solvent }} $

$ \frac{\mathrm{P}^0-\mathrm{P}^0 / 2}{\mathrm{P}^0 / 2}=\frac{{ }^{\mathrm{n}} \text { solute }}{{ }^n \text { solvent }} $

${ }^n$ solute $={ }^n$ solvent $=\frac{100}{18}=5.55 \mathrm{~mol}$

Molality of a solution of non volatile solute $(A)=1$

Elevation in boiling point is given by

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{K}_{\mathrm{b}} \mathrm{m}$

$ 3=\mathrm{K}_{\mathrm{b}} \times 1 $       ... (1)

Molality of $(A)$ in the same solvent $=2$

Depression in freezing point is given by

$ \begin{aligned} &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \\ &6=\mathrm{K}_{\mathrm{f}} \times 2 \quad\quad\quad{...(2)} \\ &\text { Dividing (1) by (2) } \\ &\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=\frac{1}{\mathrm{X}}=\frac{1}{1} \\ &\therefore \quad \mathrm{X}=1 \end{aligned} $

$\begin{aligned} &\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{iK}_{\mathrm{b}} \mathrm{m} \\\\ &\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{iK}_{\mathrm{f}} \mathrm{m} \\\\ &\frac{4}{4}=\frac{\mathrm{K}_{\mathrm{b}} 1.5}{\mathrm{~K}_{\mathrm{f}} 4.5} \\\\ &\frac{\mathrm{K}_{\mathrm{b}}}{\mathrm{K}_{\mathrm{f}}}=3 \end{aligned}$

$\Delta \mathrm{T}_{\mathrm{b}}=\mathrm{i} \times \mathrm{K}_{\mathrm{b}} \times \mathrm{m}$

Moles of solute $($ acetic acid $)=\frac{1.2 \times 1.02}{60}$

As moles of solute are very less.

So, take molarity and molality the same.

$0.0198=\mathrm{i} \times 1.85 \times \frac{1.2 \times 1.02}{60 \times 2}$

$\mathrm{i}=1.05$

$\alpha=\frac{i-1}{n-1}=\frac{0.05}{1}= 0.05 = 5\text{%}$

Since,

$ \pi=\mathrm{icR} \mathrm{T} $

$5.03 \times 10^{-3}=\frac{2.5}{M} \times \frac{1000}{500} \times 0.083 \times 300$

Molar mass of protein $=24751.5 \mathrm{~g} / \mathrm{mol}$

Number of glycine units in protein $=\frac{24751.5}{75}$

$ =330 $

$ \begin{aligned} &\frac{\mathrm{y}_{\mathrm{B}}}{1-\mathrm{y}_{\mathrm{B}}}=\frac{\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}}{\mathrm{P}_{\mathrm{A}}^{\mathrm{o}}}\left[\frac{\mathrm{X}_{\mathrm{B}}}{1-\mathrm{X}_{\mathrm{B}}}\right] \\\\ &\Rightarrow \frac{\mathrm{y}_{\mathrm{B}}}{1-\mathrm{y}_{\mathrm{B}}}=\frac{100}{50}\left[\frac{0.7}{0.3}\right]=\frac{14}{3} \\\\ &\Rightarrow \mathrm{y}_{\mathrm{B}}=\frac{14}{17} \end{aligned} $

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

$ \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 $=1221 \times 10^{-3} \mathrm{~g}$