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.

Moles of water $n_{\text {water }}=\frac{1000 \mathrm{~g}}{18 \mathrm{~g} / \mathrm{mol}}$

$ =55.56 \mathrm{~mol} $

Mole of solute $=n_{\text {solute }}=1 \mathrm{~mol}$

Mole fraction of water

$ \begin{aligned} & \chi_{\text {water }}=\frac{n_{\text {water }}}{n_{\text {water }}+n_{\text {solute }}} \\ & \chi_{\text {water }}=\frac{55.56}{55.56+1} \approx 0.9823 \end{aligned} $

Vapour pressure of solution,

$ \begin{aligned} p_{\text {solution }} & =\chi_{\text {Water }} \times p_{\text {pure water }} \\ & =0.982 \times x k P_a=0.982 x \end{aligned} $

Molality of $X Y$ and $X Y_3$

$ \begin{gathered} m_{X Y}=\frac{\Delta T_{\mathrm{PXY}}}{k_f}=\frac{5.333}{2}=2.6665 \mathrm{~mol} / \mathrm{kg} \\ m_{X Y y_3}=\frac{\Delta T_{f\left(X Y_3\right)}}{k_f}=\frac{2.2857}{2} \\ =1.14285 \mathrm{~mol} / \mathrm{kg} \end{gathered} $

Mass of solvent $=50 \mathrm{~g}=0.05 \mathrm{~kg}$

Moles of $X Y: n_{X Y}=m_{X Y} \times$ mass of solvent

$ =0.133325 \mathrm{~mol} $

Moles of $X Y_3$ :

$ n_{X Y_3}=1.14285 \times 0.05=0.05714 \mathrm{~mol} $

Molar mass of $X Y=\frac{10}{0.133325}=75 \mathrm{~g} / \mathrm{mol}$

Molar mass of $X Y_3=\frac{10}{0.0571425}$

$ =175 \mathrm{~g} / \mathrm{mol} $

Let $A_X$ be atomic weight of $X$ and $A_Y$ be atomic weight of $Y$.

$ \begin{gathered} A_X+A_Y=75 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(i)\\ A_X+3 A_Y=175 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,..(ii)\end{gathered} $

Solving Eqs. (i) and (ii),

$ A_Y=25 \mathrm{~g} / \mathrm{mol}, A_Y=50 \mathrm{~g} / \mathrm{mol} $

or 25 u and 50 u

$ \begin{aligned} &\text { Moles of ethylene glycol }\\ &\begin{aligned} & \qquad=\frac{\text { Given mass }}{\text { Molar mass }}=\frac{124 \mathrm{~g}}{62 \mathrm{~g} / \mathrm{mol}}=2 \mathrm{~mol} \\ & \text { Molality }=\frac{\text { Moles of solute }}{\text { Mass of solvent }(\mathrm{kg})} \\ & \text { Mass of water }=\frac{2 \mathrm{~mol}^{10} \mathrm{~kg} \mathrm{~mol}^{-1}}{=0.2 \mathrm{~kg} \text { or } 200 \mathrm{~g}} \end{aligned} \end{aligned} $

For an ideal gas, $x$-axis represent mole fraction and $y$ represent vapour pressure.

According to Raoult's law, vapour pressure is directly proportional to mole fraction of solvent, Since, solute is non-volatile, it does not contribute to vapour pressure.

A $10 \%(w / w)$ solution $=10 \mathrm{~g} 0 \mathrm{f}$ glucose in 100 g solution.

mass of solute $=10 \mathrm{~g}$, mass of solvent $=100-10=90 \mathrm{~g}$

$ \begin{aligned} & \text { Moles of glucose }=\frac{\text { Mass of glucose }}{\text { Molar mass }} \\ & \quad \begin{aligned} & =\frac{10}{180}=0.0556 \mathrm{~mol} \\ \text { Molality } & =\frac{\text { No. of moles of solute }}{\text { Mass of water }(\mathrm{kg})} \\ & =\frac{0.0556}{0.090} \\ & =0.62 \mathrm{~m} \end{aligned} \end{aligned} $

$ \text{Given:} $

$ i = 1.075, \quad m = 0.5 \, \text{mol kg}^{-1}, \quad K_f = 1.86 \, \text{K·kg·mol}^{-1} $

We are asked to find the experimentally observed freezing point depression, $ \Delta T_f $.

Formula

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

Substitute the values

$ \Delta T_f = 1.075 \times 1.86 \times 0.5 $

Perform the calculation

$ 1.86 \times 0.5 = 0.93 $

$ 0.93 \times 1.075 = 0.99975 \approx 1.00 $

✅ Therefore,

$ \boxed{\Delta T_f = 1.0 \, \text{K}} $

Correct Option:

Option C – 1.0

Step 1: Find the decrease in freezing point

The difference between the normal freezing point of water and the solution’s freezing point is called the depression in freezing point. The calculation is:

$ \Delta T_f = T_f^0 - T_f = 273.16 - 272.814 $

$ \Delta T_f = 0.346~\mathrm{K} $

Step 2: Use the freezing point depression formula

The formula is: $\Delta T_f = K_f \cdot m$ where $m$ is molality.

$ m = \frac{\Delta T_f}{K_f} = \frac{0.346}{1.86} = 0.186~\mathrm{mol/kg} $

Step 3: Calculate the moles of solute

Molality tells us moles of solute in 1 kg of solvent (water). Here, amount of water is $21.5~\mathrm{g} = 0.0215~\mathrm{kg}$.

Moles of solute $= 0.186 \times 0.0215 = 0.00400~\mathrm{mol}$

Step 4: Calculate the molar mass

Molar mass $= \frac{\text{mass of solute}}{\text{moles of solute}}$

$ \text{Molar mass} = \frac{0.2}{0.00400} = 50~\mathrm{g/mol} $

Mole fraction of solvent

$ \begin{aligned} \chi_{\text {solute }}= & \frac{p_0-p_s}{p_0} \\ \Rightarrow \quad \chi_{\text {Solute }}= & \frac{12.3-12.078}{12.3}=0.01805 \end{aligned} $

Moles of water in 1000 g of water

$ =\frac{1000}{18}=55.56 \mathrm{~mol} $

For dilute solution,

$ \chi_{\text {Solute }}=\frac{n_{\text {solute }}}{n_{\text {solute }}+n_{\text {water }}} \approx \frac{n_{\text {solute }}}{n_{\text {water }}} $

Since molality $x$ is moles of solute per 1 kg of solvent

$ \begin{aligned} & x=n_{\text {solute }} \Rightarrow \quad x_{\text {solute }}=\frac{x}{55.56} \\ & x=0.01805 \times 55.56 \approx 1.018 \end{aligned} $

Mole fraction of binary solution,

$ \begin{aligned} & \chi_{\text {glucose }}+\chi_{\text {water }}=1 \\ & \chi_{\text {glucose }}+40 \times \chi_{\text {glucose }}=1 \\ & \chi_{\text {glucose }}=\frac{1}{41} \end{aligned} $

Assume 1 mole of total solution Moles of glucose

$ \left(n_{\text {glucose }}\right)=\chi_{\text {glucose }} \times 1=\frac{1}{41} \mathrm{~mol} $

Moles of water $\left(n_{\text {water }}\right)=\frac{40}{41} \mathrm{~mol}$

Mass of glucose ( $M_{\text {glucose }}$ )

$ \begin{aligned} & =n_{\text {glucose }} \times \text { Molar mass } \\ & =\frac{1}{41} \times 180=\frac{180}{41} \end{aligned} $

Mass of water $\left(m_{\text {water }}\right)=n_{\text {water }} \times$ Molar mass

$ =\frac{40}{41} \times 18=\frac{720}{41} $

$w / w \%$ of glucose

$ \begin{aligned} & \frac{\text { Mass of glucose }}{\text { Total mass }} \times 100 \\ & =\text { Total mass }=\frac{180}{41}+\frac{720}{41}=\frac{900}{41} \mathrm{~g} \end{aligned} $

Weight \% of glucose

$ =\frac{\frac{180}{41}}{\frac{900}{41}} \times 100 \%=20 \% $

$ \text { } \begin{aligned} \text { Moles of benzoic acid } & =\frac{2.44}{122} \\ & =0.02 \mathrm{~mol} \end{aligned} $

$ \text { Mass of benzene }=\frac{30}{1000 \mathrm{~g} / \mathrm{kg}}=0.03 \mathrm{~kg} $

Molality, $m=\frac{0.02}{0.03}=0.6667 \mathrm{~mol} / \mathrm{kg}$

Depression in freezing point,

$ \begin{aligned} & \Delta T_f=i K_f \cdot m \\ & 2=i \times 5 \times 0.6667=0.6 \end{aligned} $

Using van't Hoff formula,

$ 0.6=1-\alpha+\frac{\alpha}{2} \Rightarrow \alpha=0.8 $

Percentage of association

$ =0.8 \times 100=80 \% $

Elevation in boiling point,

$ \Delta T_b=100.18^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}=0.18^{\circ} \mathrm{C} $

Formula for elevation in boiling point is

$ \begin{aligned} \Delta T_b & =K_b \cdot m \\ m & =\frac{0.18}{0.52}=0.346 \mathrm{~mol} / \mathrm{kg} \end{aligned} $

Depression in freezing point is,

$ \Delta T_f=K_f \cdot m=1.86 \times 0.346=0.643 \mathrm{~K} $

Freezing point of sol

$ 0^{\circ} \mathrm{C}-0.643^{\circ} \mathrm{C}=-0.64^{\circ} \mathrm{C} $

Moles of benzene

$ =n_B=\frac{23.4}{78}=0.3 \mathrm{~mol} $

Moles of toluene $=n_{\text {rolu }}=\frac{64.4}{92}=0.7 \mathrm{~mol}$ Moles fraction, $\chi_{\text {Tolu }}=\frac{n_{\text {Tolu }}}{n_{\text {Total }}}=\frac{0.7}{1}=0.7$;

$ \chi_B=1-0.7=0.3 $

Partial pressure of toluene

$ \begin{aligned} & p_{\text {Tolu }}=\chi_{\text {Total }} \cdot p_{\text {Tolu }}^{\circ} \\ \Rightarrow \quad & 0.7 \times 22=15.4 \mathrm{mmHg} \end{aligned} $

Partial pressure of benzene

$ \begin{aligned} & p_{\text {Ben }}=\chi_B \cdot p_B \\ \Rightarrow \quad & 0.3 \times 75=225 \mathrm{mmHg} \end{aligned} $

Total vapour, pressure

$ =22.5+15.4=37.9 \mathrm{mmHg} $

Mole fraction of toluene in vapour

$ Y_{\text {toluene }}=\frac{p_{\text {Toluene }}}{p_{\text {Total }}}=\frac{15.4}{37.9}=0.406 $

Depression in freezing point is given by

$ \begin{aligned} \Delta T_f & =K_f \cdot m \\ \Rightarrow \quad m & =\frac{\Delta T_f}{K_f}=\frac{0.64}{5.12}=0.125 \mathrm{~mol} / \mathrm{kg} \end{aligned} $

Mole of solute $=$ molality $\times$ mass of solvent

$ \begin{aligned} & =0.125 \times \frac{100}{1000}=0.0125 \mathrm{~mol} \\ \text { Molar mass } & =\frac{\text { Mass of solute }}{\text { Moles of solute }}=\frac{1.95}{0.0125}=156 \mathrm{~g} / \mathrm{mol} \end{aligned} $

Given data

• Moles of $ A = 0.714 \, \text{mol} $

• Moles of $ B = 5.555 \, \text{mol} $

• $ P_{\text{solution}} = 475 \, \text{torr} $

• $ P_A^0 = 280.7 \, \text{torr} $

• $ P_B^0 = ? $

Step 1. Mole fractions

$ x_A = \frac{0.714}{0.714 + 5.555} = \frac{0.714}{6.269} = 0.1138 $

$ x_B = 1 - x_A = 0.8862 $

Step 2. Raoult’s law

For an ideal solution:

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

We know everything except $ P_B^0 $.

So:

$ 475 = (0.1138)(280.7) + (0.8862) P_B^0 $

Step 3. Solve for $ P_B^0 $

Compute first term:

$ 0.1138 \times 280.7 = 31.93 $

Thus

$ 475 = 31.93 + 0.8862 P_B^0 $

$ 0.8862 P_B^0 = 475 - 31.93 = 443.07 $

$ P_B^0 = \frac{443.07}{0.8862} = 500.1 $

✅ Answer: $ P_B^0 = 500 \, \text{torr} $

Correct Option: D (500 torr)

Let’s assume we have $1$ mole of the total solution.

Mole fraction of glucose is $0.0244$. This means there are $0.0244$ moles of glucose in the solution.

The mass of glucose is calculated by multiplying the number of moles by the molar mass of glucose ($180.16\, \mathrm{g/mol}$):

$ 0.0244 \times 180.16 = 4.3959~\mathrm{g} $

The mole fraction of water is $0.9756$. This means there are $0.9756$ moles of water in the solution.

The mass of water is found by multiplying the number of moles by the molar mass of water ($18.01\, \mathrm{g/mol}$):

$ 0.9756 \times 18.01 = 17.570~\mathrm{g} $

The total mass of the solution is just the mass of glucose plus the mass of water:

$ 4.3959 + 17.570 = 21.9659~\mathrm{g} $

To find the weight percent $(w/w)$ of glucose, divide the mass of glucose by the total mass and multiply by $100$:

$ \frac{4.3959}{21.9659} \times 100 = 20~\% $

Step 1: Find the mole fraction of water.

The solution has a mole fraction of solute = 0.02.

Mole fraction of water = 1 minus mole fraction of solute.

So, $ \chi_{\text{water}} = 1 - 0.02 = 0.98 $

Step 2: Use Raoult's Law to relate vapour pressures.

Raoult’s law says: Vapour pressure of solution = Vapour pressure of pure water × Mole fraction of water.

$ p_{\text{solution}} = p_{\text{water}} \times \chi_{\text{water}} $

We know $ p_{\text{solution}} = 34.65\ \textrm{mm Hg} $ and $ \chi_{\text{water}} = 0.98 $.

So, $ 34.65 = p_{\text{water}} \times 0.98 $

Step 3: Solve for the vapour pressure of pure water.

Divide both sides by 0.98:

$ p_{\text{water}} = \frac{34.65}{0.98} = 35.36\ \textrm{mm Hg} $

Given, $T=27^{\circ} \mathrm{C}$,

$ \begin{gathered} R=0.083 \mathrm{~L} \text { at } \mathrm{mK}^{-1} \mathrm{~mol}^{-1} \\ \text { dissociation } \%=50 \% \\ \pi=? \end{gathered} $

The dissociation of acetic acid

$ \begin{gathered} \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+} \\ n=2 \end{gathered} $

Degree of dissociation $=50 \%$ or 0.5

$ i=1+0.5(2-1)=1.5 $

Osmotic pressure is given by

$ \begin{aligned} \pi & =i C R T \\ & =1.5 \times 0.01 \times 0.023 \times 300=0.37 \mathrm{~atm} \end{aligned} $

Given data

• $ P_A^0 = 70 \ \text{mm Hg} $

• Ideal solution → obeys Raoult's law

• $ x_B = 0.2 $ ⇒ $ x_A = 1 - 0.2 = 0.8 $

• Total vapour pressure of solution $ P_\text{total} = 84 \ \text{mm Hg} $

• Temperature = 300 K (constant)

We are asked to find $ P_B^0 $ (vapour pressure of pure liquid $B$).

Step 1: Apply Raoult’s Law

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

Substitute known values:

$ 84 = (0.8)(70) + (0.2)(P_B^0) $

Step 2: Simplify

$ 84 = 56 + 0.2 P_B^0 $

$ 84 - 56 = 0.2 P_B^0 $

$ 28 = 0.2 P_B^0 $

Step 3: Solve for $P_B^0$

$ P_B^0 = \frac{28}{0.2} = 140 \ \text{mm Hg} $

✅ Answer: $ P_B^0 = 140\ \text{mm Hg}$

Correct option: A (140)

$ \begin{aligned} &\text { Moles (n) of ethylene glycol, }\\ &\begin{aligned} & \quad=\frac{\text { Mass }}{\text { Molar mass }} \\ & \Rightarrow \quad n=\frac{248}{62}=4 \mathrm{~mol} \\ & \text { Mass of water (in kg) }=0.2 \mathrm{~kg} \\ & \text { Molality }=\frac{\text { Moles of solute }}{\text { Mass (in kg) of solvent }} \\ & \qquad=\frac{4 \mathrm{~mol}}{0.2}=20 \mathrm{~mol} / \mathrm{kg} \end{aligned} \end{aligned} $

Moles of urea

$ =\frac{\text { Mass }}{\text { Molar mass }}=\frac{7.5}{60}=0.125 \mathrm{~mol} $

$ \begin{aligned} \text { Molality of urea } & =\frac{\text { Moles of urea }}{\text { Mass of water in } \mathrm{kg}} \\ & =\frac{0.125}{1}=0.125 \mathrm{~mol} / \mathrm{kg} \end{aligned} $

Since both solution freezes at the same temperature, their molalities are equal.

$ \text { } \begin{aligned}\text { Molality of } x & =\text { Molality of urea } \\ & =0.125 \mathrm{~mol} / \mathrm{kg} \end{aligned} $

Moles of $x=$ Molality $\times$ mass of water

$ =0.125 \times 1 $

$\Rightarrow \quad 0.125 \mathrm{~mol}$

$ \begin{aligned} \text { Molar mass of } x & =\frac{\text { Mass }}{\text { Moles }}=\frac{15}{0.125} \\ & =120 \mathrm{~g} / \mathrm{mol} \end{aligned} $

Magnesium nitrate separates into ions in solution as shown below:

$ \mathrm{Mg}\left(\mathrm{NO}_3\right)_2 \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{NO}_3^{-} $

This means that one unit of magnesium nitrate produces a total of 3 particles after complete dissociation (1 magnesium ion and 2 nitrate ions).

If the salt is only 40% dissociated, it means $\alpha = 0.4$.

The Van’t Hoff factor, $ i $, which shows how many times the number of effective particles increases due to dissociation, is given by:

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

Substituting the values, we get:

$ i = 1 + 0.4(3 - 1) = 1 + 0.8 = 1.8 $

Elevation of boiling point depends on the vant' Hoff factor (i).

Urea is non-electrolyte, $i=1$

$\mathrm{CaCl}_2$ dissociates into 3 ions, $i=3$

So, effective particles $=0.05 \times 3=0.15 \mathrm{~mol}$ for $\mathrm{CaCl}_2$ is more than 0.1 mol for urea is equal amount of solvent.

Hence, $\mathrm{CaCl}_2$ solution will show greater elevation.

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} $

Given the vapor pressures of liquids $ A $ and $ B $ at 300 K are $ p_A^{\circ} = 500 \, \text{mmHg} $ and $ p_B^{\circ} = 400 \, \text{mmHg} $, respectively, and knowing that equal moles of $ A $ and $ B $ are mixed to form an ideal solution, we proceed as follows:

Since equal moles of $ A $ and $ B $ are mixed:

$ X_A = X_B = 0.5 $

Using Raoult's law, we calculate the partial pressures:

For $ A $:

$ p_A = X_A \times p_A^{\circ} = 0.5 \times 500 = 250 \, \text{mmHg} $

For $ B $:

$ p_B = X_B \times p_B^{\circ} = 0.5 \times 400 = 200 \, \text{mmHg} $

Total pressure of the solution:

$ \text{Total pressure} = p_A + p_B = 250 + 200 = 450 \, \text{mmHg} $

Mole fraction of $ A $ in the vapor phase:

$ Y_A = \frac{p_A}{\text{Total pressure}} = \frac{250}{450} = 0.555 $

Mole fraction of $ B $ in the vapor phase:

$ Y_B = \frac{p_B}{\text{Total pressure}} = \frac{200}{450} = 0.444 $

Both statements are correct.

Statement I: Positive deviation from Raoult’s law means A–B interactions are weaker than A–A and B–B, so this is true.

Statement II: An ideal solution has no heat or volume change on mixing, i.e.

$\Delta_{\rm mix}H = 0,\quad \Delta_{\rm mix}V = 0\,,$

so this is also true.

Answer: Option A.

At 290 K, vessel (I) contains equal moles of liquids A, B, and C. Their boiling points are :

A: 350 K

B: 373 K

C: 308 K

Since a lower boiling point indicates a higher volatility (and hence a higher vapor pressure at a given temperature), at 290 K the order of vapor pressures will be:

$ p_C > p_A > p_B $

This confirms statement (III).

Now, vessel (I) is heated to 300 K. At this higher temperature:

Liquid C (boiling point 308 K) becomes much more volatile.

Liquid A (boiling point 350 K) is moderately volatile.

Liquid B (boiling point 373 K) is the least volatile.

As a result, the more volatile liquids (C and A) evaporate preferentially. This means:

The remaining liquid in vessel (I) will be enriched in B because it evaporates very little. So, statement (I) "Vessel - I is rich in liquid B" is correct.

The vapours collected in vessel (II) will be richer in C, since C evaporates the most. So, statement (II) "Vessel - II is rich in vapour of C" is also correct.

In summary, all three statements are correct:

Statement (I): True.

Statement (II): True.

Statement (III): True.

Thus, the correct answer is Option A: I, II, III.

To calculate the osmotic pressure of a solution where 6 grams of urea is dissolved in 500 mL of water at 300 K, we'll use the following formula for osmotic pressure:

$ \pi = \frac{n}{V} \times R \times T $

Given:

$ R = 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} $

Temperature, $ T = 300 \, \text{K} $

Volume of solution, $ V = 500 \, \text{mL} = 0.5 \, \text{L} $

First, we need to determine the number of moles of urea ($ n $). The molar mass of urea ($\text{NH}_2\text{CONH}_2$) is calculated as follows:

$ \text{Molar mass of urea} = 1 \times 12 + 4 \times 1 + 2 \times 14 + 1 \times 16 = 60 \, \text{g/mol} $

Therefore, the number of moles, $ n $, is:

$ n = \frac{\text{Mass}}{\text{Molar mass}} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{mol} $

Substitute back into the osmotic pressure formula:

$ \pi = \frac{n}{V} \times R \times T = \frac{0.1 \, \text{mol}}{0.5 \, \text{L}} \times 0.082 \, \text{L atm K}^{-1} \text{mol}^{-1} \times 300 \, \text{K} $

Calculate the values:

$ \pi = \frac{0.1}{0.5} \times 0.082 \times 300 = 4.92 \, \text{atm} $

Thus, the osmotic pressure of the solution is 4.92 atm.

In the context of Henry's law, the relationship between the Henry's constant ($K_{\mathrm{H}}$) and solubility ($S$) can be understood as follows:

$ K_{\mathrm{H}} \propto \frac{1}{S} $

This equation indicates that the Henry's constant is inversely proportional to the solubility of a gas in a liquid, given a specific external pressure and temperature. In simple terms, if a gas is less soluble in water, its Henry's constant will be higher.

For helium (He), it is known to have the lowest solubility among the gases considered here. Therefore, helium will have the highest Henry's constant at 293 K.

Given:

Weight of benzene (solvent), $ W_A = 49 \, \text{g} = 49 \times 10^{-3} \, \text{kg} $

Depression in freezing point, $ \Delta T_f = 1.12 \, \text{K} $

Cryoscopic constant for benzene, $ K_f = 4.9 \, \text{K} \, \text{kg/mol} $

Molar mass of benzoic acid, $ M_B = 122 \, \text{g/mol} $

Degree of association, $ \alpha = 0.88 $

Formula for depression in freezing point:

$ \Delta T_f = \frac{i \times K_f \times W_B}{M_B \times W_A} $

Calculating the van't Hoff factor ($ i $) for association:

For dimerization (association of 2 molecules into 1 dimer), the expression for $ i $ is:

$ i = 1 - \left(1 - \frac{1}{n}\right) \times \alpha $

Substituting the values:

$ i = 1 - \left(1 - \frac{1}{2}\right) \times 0.88 = 0.56 $

Solving for $ W_B $ (weight of solute, $ x $):

$ W_B = \frac{\Delta T_f \times W_A \times M_B}{i \times K_f} $

Substituting the known values:

$ W_B = \frac{1.12 \times 49 \times 10^{-3} \times 122}{0.56 \times 4.9} $

Calculating:

$ W_B = 2.44 \, \text{g} $

Therefore, the weight of benzoic acid, $ x $, is $ 2.44 \, \text{g} $.

Given Data

Liquid $ A $

Pure vapor pressure, $ p_A^{\circ} = 400 $ mm Hg

Mole fraction, $ X_A = 0.7 $ (since $ X_B = 0.3 $)

Liquid $ B $

Pure vapor pressure, $ p_B^{\circ} = 600 $ mm Hg

Mole fraction, $ X_B = 0.3 $

Total Vapor Pressure Calculation

According to Dalton’s law of partial pressures, the total vapor pressure ($ p_{\text{total}} $) of the solution is given by:

$ p_{\text{total}} = X_A \cdot p_A^{\circ} + X_B \cdot p_B^{\circ} $

Substituting the values:

$ p_{\text{total}} = 0.7 \times 400 + 0.3 \times 600 = 280 + 180 = 460 \text{ mm Hg} $

Mole Fraction in Vapor Phase

Mole Fraction of $ A $, $ Y_A $

The mole fraction of $ A $ in the vapor phase is calculated as follows:

$ Y_A = \frac{p_A}{p_{\text{total}}} = \frac{0.7 \times 400}{460} = \frac{280}{460} = 0.609 $

Mole Fraction of $ B $, $ Y_B $

The mole fraction of $ B $ in the vapor phase can be found by:

$ Y_B = 1 - Y_A = 1 - 0.609 = 0.391 $

Therefore, the mole fractions of $ A $ and $ B $ in the vapor phase are 0.609 and 0.391, respectively.

Water boils at $100^{\circ} \mathrm{C}$ or 373.13 K becomes at this temperature, the vapour pressure of waer is equal to the atmospheric pressure.

However, vapour pressure of liquid decreases in presence of non-volatile solution like urea.

In order that the vapour pressure reaches the atomospheric pressure, the temperature is increase beyond normal boiling point of pure water.

Hence, urea solution temperature (boiling) is always higher than pure solvent (water).

$ \text { At } 1 \text { bar, boiling point of urea is } T_3 \text {. } $

Statement I is incorrect, while Statement II is correct. The corrected version of Statement I should state that for a non-ideal solution with negative deviation, the interactions between different components ($A-B$) are stronger than the interactions within the same components ($A-A$ and $B-B$). This results in a negative change in volume and enthalpy upon mixing ($\Delta V_{\text{mixing}} = -\text{ve}$ and $\Delta H_{\text{mixing}} = -\text{ve}$).

Statement II asserts that in reverse osmosis, the applied pressure must exceed the osmotic pressure of the solution for the process to occur, which is correct.

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

The formula for calculating the relative lowering of vapor pressure is:

$ \frac{p_0-p}{p_0} = \frac{x_A}{x_A + x_B} $

Where:

$x_A$ is the moles of the solute.

$x_B$ is the moles of the solvent.

For this solution, we have:

$x_A = 0.1$ moles (solute)

$x_B = 0.9$ moles (solvent)

Plugging these values into the formula, we get:

$ \frac{p_0-p}{p_0} = \frac{0.1}{0.1+0.9} = 0.1 $

Therefore, the relative lowering of vapor pressure is 0.1.

To determine the freezing point of the solution, we use the concept of colligative properties. The elevation in boiling point ($\Delta T_b$) is given as 0.052 K, with the elevation constant ($K_b$) for water at 0.52 K kg/mol. The freezing point depression constant ($K_f$) is 1.86 K kg/mol, and the original freezing point of water is 273 K.

First, calculate the molality ($m$) from the elevation in boiling point:

$ \Delta T_b = m \times K_b \quad \Rightarrow \quad m = \frac{\Delta T_b}{K_b} $

$ m = \frac{0.052 \, \text{K}}{0.52 \, \text{K} \, \text{kg/mol}} = 0.1 $

Next, use this molality to find the depression in freezing point ($\Delta T_f$):

$ \Delta T_f = m \times K_f = 0.1 \times 1.86 = 0.186 \, \text{K} $

Finally, calculate the freezing point of the solution ($T_f$):

$ T_f = 273 \, \text{K} - \Delta T_f = 273 \, \text{K} - 0.186 \, \text{K} = 272.814 \, \text{K} $

Thus, the freezing point of the solution is 272.814 K.

From experiment I and III pure water can be obtained in vessel II. This is due to phenomenon of reverse osmosis.

Reverse osmosis is the process of movement of solvent through a SPM from the solution to the pure solvent by applying excess pressure ( $>\pi$ ) on solution side (in this case sea water.)

To find the van't Hoff factor for the KCl solution, we start with the given data:

Elevation in boiling point, $\Delta T_b = 0.01 \, \text{K}$

Molal boiling point elevation constant for water, $K_b = 0.52 \, \text{K} \, \text{kg} \, \text{mol}^{-1}$

Molality of the solution, $m = 0.01 \, \text{m}$

The formula for the elevation in boiling point is:

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

Where $i$ is the van't Hoff factor.

Rearranging the formula to solve for $i$, we get:

$ i = \frac{\Delta T_b}{K_b \times m} $

Substituting the given values:

$ i = \frac{0.01}{0.52 \times 0.01} $

Calculating the result:

$ i = 1.92 $

Let's break down the problem step by step:

Calculate the mass of urea in the first solution:

The first solution weighs 200 g and is 20% urea.

Mass of urea = $200 \, \text{g} \times 0.20 = 40 \, \text{g}$.

Calculate the mass of urea in the second solution:

The second solution weighs 400 g and is 40% urea.

Mass of urea = $400 \, \text{g} \times 0.40 = 160 \, \text{g}$.

Determine the total mass of urea and the total mass of the solution:

Total urea = $40 \, \text{g} + 160 \, \text{g} = 200 \, \text{g}$.

Total solution mass = $200 \, \text{g} + 400 \, \text{g} = 600 \, \text{g}$.

Find the weight percentage of urea in the resultant solution:

Weight percentage =

$\frac{\text{Total mass of urea}}{\text{Total mass of solution}} \times 100\% = \frac{200 \, \text{g}}{600 \, \text{g}} \times 100\% \approx 33.33\%$.

Thus, the weight percentage of urea in the final solution is approximately 33.33%, which corresponds to Option B.

To determine the number of moles of methane dissolved in 1 L of water at 293 K with a partial pressure of 1 bar, we employ Henry's law and the concept of mole fraction.

Given:

Partial pressure of methane, $ p_0 = 1 $ bar

Henry's constant for methane, $ K_{\mathrm{H}} = 0.4 \, \mathrm{K} \, \text{bar} $

Step 1: Apply Henry's Law

Henry's law can be expressed as:

$ p_0 = K_{\mathrm{H}} \chi $

Solving for the mole fraction ($\chi$) of methane, we have:

$ 1 = 0.4 \times 10^3 \times \chi $

$ \chi = 2.5 \times 10^{-3} \tag{Equation 1} $

Step 2: Calculate Mole Fraction

The mole fraction of methane ($x$) in the solution is given by:

$ x = \frac{n (\text{methane})}{n (\text{methane}) + n (\text{water})} \tag{Equation 2} $

With 1 L of water, the number of moles of water is:

$ n (\text{water}) = \frac{1000 \, \text{g}}{18 \, \text{g/mol}} = 55.5 \, \text{mol} $

Assuming $ n (\text{methane}) $ is significantly smaller than 55.5 mol, substituting into Equation 2 simplifies to:

$ 2.5 \times 10^{-3} = \frac{n}{55.5} $

Solving for $ n $, the moles of methane:

$ n = 2.5 \times 10^{-3} \times 55.5 $

$ n = 1.38 \times 10^{-1} $

Thus, the number of moles of methane dissolved in 1 L of water is $ 1.38 \times 10^{-1} $.