Q.2 g of benzoic acid ($C_6H_5COOH$) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol$^{-1}$. What is the percentage association of acid if it forms dimer in solution?

Concept & Intuition

Raoult’s law for a non‑volatile solute tells us that the relative lowering of vapour pressure is exactly equal to the mole fraction of the solute in the solution:

Here the given value is 0.0125, so $x_{\text{solute}} = 0.0125$.

We need molality (moles of solute per kg of solvent). For a dilute aqueous solution, the mole fraction of solute is small, so the mole fraction of water is nearly 1. This lets us convert directly:

since $n_{\text{solute}} \ll n_{\text{water}}$. Then molality $m = \frac{n_{\text{solute}}}{\text{kg of water}}$. For water, 1 kg contains $1000/18 \approx 55.56$ moles. So $n_{\text{water}} \approx 55.56$ per kg. Thus:

Step‑by‑step calculation

1. Write the exact relation

2. Express in terms of molality Let $m$ be molality (mol solute per kg water). For 1 kg water, $n_{\text{water}} = \frac{1000}{18} = 55.555\ldots$ mol. Then $n_{\text{solute}} = m$. So:

3. Solve for $m$ Multiply both sides:

4. Round to two decimal places $m \approx 0.70$ m.

The key idea here is that enthalpy of combustion is just a special case of enthalpy of reaction. For any reaction,

where $\Delta H_f^\circ$ is the standard enthalpy of formation. Combustion means burning in excess oxygen to give CO₂(g) and H₂O(l) as the only products. So we first write the balanced combustion equation for benzoic acid, then plug in the given formation enthalpies.

1. Write the balanced combustion reaction Benzoic acid is C₆H₅COOH. Its molecular formula is C₇H₆O₂. Complete combustion with O₂ gives CO₂ and H₂O:

Check: left has 7 C, 6 H, 2 + 15 = 17 O; right has 7×2 + 3×1 = 17 O. Balanced.

2. Apply Hess’s law The enthalpy of combustion is $\Delta H_{\text{comb}} = \Delta H_{\text{reaction}}$ for this equation. Using formation enthalpies:

Note: $\Delta H_f^\circ$ for any element in its standard state (like O₂ gas) is zero by definition. So the O₂ term vanishes.

3. Substitute the given values $\Delta H_f^\circ(\text{CO}_2) = -393$ kJ/mol, $\Delta H_f^\circ(\text{H}_2\text{O}) = -286$ kJ/mol, $\Delta H_f^\circ(\text{C}_7\text{H}_6\text{O}_2) = -409$ kJ/mol.

Compute stepwise:

$7 \times (-393) = -2751$

$3 \times (-286) = -858$

Sum of products = $-2751 - 858 = -3609$ kJ

Subtract reactant: $-3609 - (-409) = -3609 + 409 = -3200$ kJ/mol.

4. Interpret the sign The negative sign confirms that combustion is exothermic — heat is released. The magnitude is 3200 kJ per mole of benzoic acid.

The problem asks how many metal atoms can be placed in a two-dimensional open box of length 35 Å and width 35 Å, given that the metal crystallizes in a bcc lattice with edge length 4 Å. The phrase "two-dimensional open box" means we are arranging atoms in a flat, square region — essentially a grid — and we need to know the spacing between atoms in the bcc structure when projected onto a plane.

In a bcc lattice, atoms touch along the body diagonal, not along the edge. The nearest neighbour distance is half the body diagonal: $\frac{\sqrt{3}}{2}a$. For $a = 4$ Å, this distance is $\frac{\sqrt{3}}{2} \times 4 = 2\sqrt{3} \approx 3.464$ Å. This is the centre-to-centre distance between adjacent atoms in the lattice. When placing atoms in a box, each atom effectively occupies a square of side equal to this distance (since we cannot overlap atoms and they must be placed at lattice positions).

Now, we simply see how many such spacings fit along each dimension of the box.

1. Find the spacing between atoms.

   In bcc, the nearest neighbour distance $d = \frac{\sqrt{3}}{2}a = 2\sqrt{3}$ Å. This is the minimum centre-to-centre separation.

2. Number of atoms along one side.

   Along a length of 35 Å, the number of atoms that can be placed is the integer part of $35 / d$, because you need one full spacing per atom (the first atom can be at one edge, then each subsequent atom requires a full $d$ distance). So

3. Same for the other side.

   The width is also 35 Å, so the same calculation gives 10 atoms along that direction.

4. Total atoms in the square box.

   The arrangement is a grid of $10 \times 10 = 100$ atoms.

Concept & Intuition

This problem is about vapour composition above an ideal liquid mixture. Raoult’s law tells us that the partial pressure of each component in the vapour equals its mole fraction in the liquid times its pure vapour pressure. The total pressure is the sum of these partial pressures. The vapour composition is then the ratio of a component’s partial pressure to the total pressure. The key insight: even if the liquid has equal amounts of two substances, the vapour will be richer in the more volatile component (the one with higher pure vapour pressure). Here, benzene is more volatile than toluene, so the vapour should contain more benzene than the liquid does.

Step-by-step solution

1. Identify the given data
   • Temperature: $T = 300\ \text{K}$
   • Pure vapour pressure of toluene: $P_\text{toluene}^\circ = 3.63\ \text{kPa}$
   • Pure vapour pressure of benzene: $P_\text{benzene}^\circ = 9.7\ \text{kPa}$
   • Mole fraction of toluene in the liquid: $x_\text{toluene} = 0.4$
   • Since it’s a binary mixture, mole fraction of benzene in the liquid:

2. Apply Raoult’s law to find partial pressures Raoult’s law: $P_i = x_i P_i^\circ$
   • Partial pressure of toluene:

• Partial pressure of benzene:

3. Calculate total vapour pressure

4. Find vapour composition The mole fraction of a component in the vapour equals its partial pressure divided by the total pressure:

So the vapour contains about 20% toluene and 80% benzene.

Concept & Intuition

When a non‑volatile solute is dissolved in a volatile solvent, the vapour pressure of the solvent above the solution is lowered. Raoult’s law states that the vapour pressure of the solvent in the solution ($p$) equals the vapour pressure of the pure solvent ($p^\circ$) multiplied by its mole fraction in the liquid phase:

Here, the solute does not contribute to the vapour, so the entire vapour pressure comes from benzene. The problem asks for the number of moles of solute per mole of benzene — that is, the ratio $n_{\text{solute}} / n_{\text{benzene}}$. If we let $n_B = 1$ mole of benzene, then the mole fraction of benzene is

and we can solve for $n_{\text{solute}}$.

Step‑by‑step solution

1. Write Raoult’s law for the solution Given: $p^\circ = 268$ torr (pure benzene vapour pressure) $p = 167$ torr (solution vapour pressure)

2. Solve for the mole fraction of benzene

Simplify the fraction:

3. Express $x_B$ in terms of moles Let $n_B = 1$ mole of benzene and let $n_s$ = moles of solute. Then

Equate to the value from step 2:

4. Solve for $n_s$ Invert both sides:

Compute the decimal:

5. Interpret the result This means that for every 1 mole of benzene, we need about 0.605 moles of non‑volatile solute to lower the vapour pressure from 268 torr to 167 torr at 50 °C.

Concept and Intuition

Water, as we commonly know it, is $H_2O$, where $H$ represents protium, the most common isotope of hydrogen with one proton and no neutrons. Heavy water, on the other hand, is $D_2O$, where $D$ represents deuterium, an isotope of hydrogen with one proton and one neutron. This makes deuterium approximately twice as heavy as protium.

This difference in mass, though seemingly small, significantly affects the physical properties of water. The heavier deuterium atoms lead to stronger intermolecular forces, specifically hydrogen bonds, in $D_2O$ compared to $H_2O$. This is primarily because the vibrational energy of the O-D bond is lower than that of the O-H bond (due to the higher reduced mass), which translates to a lower zero-point energy. A lower zero-point energy means the bond is effectively stronger and requires more energy to break. Consequently, more thermal energy is needed to disrupt the ordered structure of ice and transition it into liquid water. This results in higher melting (freezing) and boiling points for heavy water.

Step-by-Step Solution

1. Recall the freezing point of normal water:

   Normal water ($H_2O$) freezes at $0~^\circ \mathrm{C}$ at $1$ atm pressure. This is a fundamental reference point for understanding the properties of water.

2. Understand the nature of heavy water ($D_2O$):

   Heavy water contains deuterium ($D$) instead of protium ($H$). Deuterium is an isotope of hydrogen with an atomic mass of approximately $2$ amu, while protium has an atomic mass of approximately $1$ amu. This difference in mass affects the physical properties of the molecule.

3. Identify the freezing point of heavy water:

   Due to the stronger hydrogen bonding and lower zero-point energy associated with the heavier deuterium atoms, heavy water requires more energy to transition from solid to liquid phase. Therefore, its freezing point is higher than that of normal water. The experimentally determined freezing point of heavy water at $1$ atm pressure is approximately $3.81~^\circ \mathrm{C}$.

   Note

   The stronger hydrogen bonds in $D_2O$ also lead to a higher boiling point ($101.4~^\circ \mathrm{C}$ for $D_2O$ compared to $100~^\circ \mathrm{C}$ for $H_2O$) and a higher density ($1.1056~\mathrm{g/cm^3}$ for $D_2O$ at $25~^\circ \mathrm{C}$ compared to $0.997~\mathrm{g/cm^3}$ for $H_2O$ at $25~^\circ \mathrm{C}$).

4. Compare with the given options:

   • (A) $0~^\circ \mathrm{C}$: This is the freezing point of normal water.
   • (B) $3.8~^\circ \mathrm{C}$: This value is very close to the actual freezing point of heavy water ($3.81~^\circ \mathrm{C}$).
   • (C) $4.8~^\circ \mathrm{C}$: This is too high.
   • (D) $1~^\circ \mathrm{C}$: This is too low.

   Therefore, option (B) is the correct answer.

The key concept here is that molarity is defined as moles of solute per litre of solution. Moles of NaOH are fixed once dissolved, but the volume of the solution changes with temperature because liquids expand when heated. At 90 °C, the solution occupies a larger volume than at 25 °C, so the molarity drops.

Let’s work through it step by step.

1. Find the moles of NaOH.

   Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol.

   Given mass = 2 g.

   Moles of NaOH = $\frac{2}{40} = 0.05$ mol.

2. Calculate molarity at 25 °C.

   Volume at 25 °C = 200 mL = 0.200 L.

   Molarity at 25 °C = $\frac{0.05}{0.200} = 0.25$ M.

3. What happens at 90 °C?

   The number of moles of NaOH stays the same — no solute is added or removed. But the solution expands. Water (and dilute aqueous solutions) has a coefficient of volume expansion of roughly $2.1 \times 10^{-4}$ °C⁻¹. Over a 65 °C rise (from 25 °C to 90 °C), the volume increases by about

   $\Delta V = V_0 \cdot \gamma \cdot \Delta T = 200 \times 2.1\times10^{-4} \times 65 \approx 2.73$ mL.

   So the new volume ≈ 202.73 mL = 0.20273 L.

4. Molarity at 90 °C = $\frac{0.05}{0.20273} \approx 0.2466$ M, which is less than 0.25 M.

The key concept here is the geometry of hydrogen bonding. For a hydrogen bond to form within a single molecule, the donor group (like –OH) and the acceptor atom (like a carbonyl oxygen) must be positioned so that a stable, usually five- or six-membered ring can close. This is only possible when the groups are on adjacent carbon atoms in a rigid structure like a benzene ring.

Let’s examine each option:

1. Phenol (A) – Phenol has an –OH group attached to a benzene ring. The only nearby acceptor atoms are the ring carbons, which are not electronegative enough to form a strong hydrogen bond. The oxygen of the –OH can act as an acceptor for another molecule, but within the same molecule, there is no suitable acceptor in the right position. So, no intramolecular H-bond.

2. Benzoic acid (B) – The –COOH group has both a donor (–OH) and an acceptor (C=O). However, these are part of the same functional group and are already bonded to each other. The carbonyl oxygen is too far from the –OH of another benzoic acid molecule to form an intramolecular bond; instead, benzoic acid forms intermolecular dimers (two molecules linked by two H-bonds). No intramolecular bond here.

3. para-Nitrophenol (C) – The –OH group is at the para position relative to the –NO₂ group. The distance between them is too large (four carbon atoms apart) to allow a stable ring. The nitro group’s oxygen is a good acceptor, but the geometry forces the H-bond to be intermolecular. So, no intramolecular bond.

4. 2-Hydroxybenzoic acid (D) – Here, the –OH group is at the ortho position relative to the –COOH group. This places the –OH hydrogen very close to the carbonyl oxygen of the –COOH. A six-membered ring (including the hydrogen) can form, which is highly stable. This is the classic example of intramolecular hydrogen bonding, often seen in salicylic acid.

The key idea here is that a weak acid does not dissociate completely. The degree of dissociation, $\alpha$, tells you what fraction of the original acid molecules have broken into ions at equilibrium. For a monoprotic acid like HA, the equilibrium constant $K_c$ (also called the acid dissociation constant $K_a$) is directly linked to $\alpha$ and the initial concentration $C$ through a simple relation — but only if we make the standard approximation that $\alpha$ is small.

Let’s walk through it.

1. Set up the equilibrium. For the reaction

start with initial concentration $C = 0.1$ M of HA, and zero of the ions. At equilibrium, a fraction $\alpha$ of HA has dissociated. So:

• $[\text{HA}] = C(1 - \alpha)$
• $[\text{H}^+] = C\alpha$
• $[\text{A}^-] = C\alpha$

2. Write the expression for $K_c$.

3. Plug in the numbers. Here $C = 0.1$ and $\alpha = 5\% = 0.05$.

4. Simplify.

Rounded to two significant figures, this is $2.6 \times 10^{-4}$.

Colligative properties — like boiling point elevation, freezing point depression, and osmotic pressure — depend only on the number of solute particles in a given amount of solvent, not on the chemical identity of those particles. This is the core idea: more particles means a larger effect.

For ionic compounds dissolved in water, the number of particles per formula unit equals the number of ions produced upon dissociation. So the key is to count ions.

1. NaCl dissociates as:

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

   That gives 2 ions per formula unit.

2. MgCl₂ dissociates as:

   $\text{MgCl}_2 \rightarrow \text{Mg}^{2+} + 2\text{Cl}^-$

   That gives 3 ions per formula unit.

3. BaCl₂ dissociates as:

   $\text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^-$

   That also gives 3 ions per formula unit.

At the same molar concentration, the solution with more ions will show a greater colligative effect. So MgCl₂ and BaCl₂, both giving 3 ions, will have equal and larger effects than NaCl, which gives only 2 ions.

Thus the correct order is:

$\text{MgCl}_2 = \text{BaCl}_2 > \text{NaCl}$