Q.On dissolving sugar in water at room temperature solution feels cool to touch. Under which of the following cases dissolution of sugar will be most rapid?

The Tyndall effect is the visible scattering of light when a beam passes through a medium containing particles large enough to scatter it — typically in the colloidal size range (1–1000 nm). This is why a projector beam cuts through fog or why a laser becomes visible in a glass of milk. The key idea is that the particles must be of a size comparable to the wavelength of visible light; if they are too small (as in a true solution), the light passes through without scattering, and the beam remains invisible from the side.

Let’s examine each option:

1. Clouds — Clouds are made of tiny water droplets or ice crystals suspended in air. These droplets are in the colloidal range, so they scatter sunlight strongly. That’s why clouds appear white (all colours scattered equally) and why you see a distinct beam of light through a gap in the clouds. Clouds definitely show the Tyndall effect.

2. Milk — Milk is a classic example of a colloid: it contains fat globules and protein micelles dispersed in water. These particles are large enough to scatter light, which is why milk looks opaque and white. A beam of light passing through milk is clearly visible from the side. Milk shows the Tyndall effect.

3. Sugar solution — When sugar dissolves in water, it breaks down into individual molecules (sucrose molecules, about 1 nm in size). This is a true solution — the solute particles are far smaller than the wavelength of light. There is no scattering; the solution is transparent and a light beam passing through it is invisible from the side. Sugar solution does not show the Tyndall effect.

4. Suspension — A suspension (like muddy water or chalk powder in water) contains large, visible particles that settle on standing. These particles are much larger than colloidal size, so they scatter light strongly — often making the mixture opaque. Suspensions do show the Tyndall effect.

The concept here is what makes a liquid mixture "ideal" in the thermodynamic sense. Unlike an ideal gas, which is about gas-phase behavior, an ideal solution is about how the components mix at the molecular level. The key condition is that the mixture obeys Raoult’s law — that is, the partial vapor pressure of each component is proportional to its mole fraction in the liquid phase. This law holds exactly only when the intermolecular forces between all pairs of molecules are identical, so that there is no net energy change or volume change upon mixing. Let’s examine each statement.

1. Statement (a): "It obeys ideal gas equation"

   This is irrelevant. The ideal gas equation ($PV = nRT$) describes the behavior of gases, not liquid mixtures. A liquid mixture is not a gas, so this condition has nothing to do with an ideal solution. Statement (a) is false.

2. Statement (b): "It obeys Raoult’s law at all concentrations"

   This is the defining criterion. Raoult’s law states that for a component $i$, $P_i = x_i P_i^*$, where $x_i$ is its mole fraction in the liquid and $P_i^*$ is its vapor pressure when pure. An ideal solution obeys this law exactly over the entire range of composition. Statement (b) is true.

3. Statement (c): "Solute – solute, solute – solvent and solvent – solvent interactions are similar"

   This is the molecular reason behind Raoult’s law. If all intermolecular forces are equal in strength, then mixing causes no net change in energy (enthalpy of mixing is zero) and no volume change (volume of mixing is zero). Under these conditions, the vapor pressure follows Raoult’s law exactly. Statement (c) is true.

Since statements (b) and (c) are correct, the answer is the option that includes both.

The key concept is that an ideal solution is one where the intermolecular forces between all molecules (solute–solute, solvent–solvent, and solute–solvent) are identical. This means that when you mix the components, there is no net energy change and no net volume change — the mixture behaves as if the molecules simply “replace” each other without any interaction effects.

Let’s examine each statement:

1. Statement a: $\Delta V_{\text{mix}} = 0$

   In an ideal solution, the volume of the mixture is exactly the sum of the volumes of the pure components before mixing. There is no contraction or expansion because the molecular packing is unchanged. This is a defining property. So a is correct.

2. Statement b: $V_{\text{solvent}} + V_{\text{solute}} = V_{\text{solution}}$

   This is simply another way of saying $\Delta V_{\text{mix}} = 0$ — the total volume after mixing equals the sum of the volumes before mixing. So b is also correct (it’s equivalent to a).

3. Statement c: $\Delta H_{\text{mix}} = 0$

   Because the intermolecular forces are all the same, no heat is absorbed or released when mixing. The enthalpy change is zero. This is the other defining property of an ideal solution. So c is correct.

4. Statement d: $\mathrm{H_2O} + \mathrm{CO_2} \rightarrow \mathrm{H_2CO_3}$ is an example of an ideal solution.

   This is a chemical reaction, not a physical mixing of components that remain as themselves. In an ideal solution, the components do not react; they simply intermingle. Moreover, $\mathrm{CO_2}$ in water does not obey Raoult’s law (it reacts and has a non-ideal behavior). So d is incorrect.

Thus, the correct statements are a, b, and c.

The key idea here is that an ideal solution is defined by having no change in enthalpy or volume upon mixing — the intermolecular forces between unlike molecules are identical to those between like molecules. So mixing is purely an entropy-driven process.

When two pure substances A and B are mixed, the molecules have more possible arrangements in the mixture than they did in the separate pure states. This increase in microstates means the entropy of the system increases: $\Delta S_{\text{sys}} > 0$.

Since $\Delta H = 0$ for ideal mixing, and the process occurs at constant temperature and pressure, the entropy change of the surroundings is given by $\Delta S_{\text{surr}} = -\frac{\Delta H}{T} = 0$. The surroundings neither gain nor lose heat.

Spontaneity at constant $T$ and $P$ is governed by the Gibbs free energy change: $\Delta G = \Delta H - T\Delta S_{\text{sys}}$. With $\Delta H = 0$ and $\Delta S_{\text{sys}} > 0$, we get $\Delta G = -T\Delta S_{\text{sys}} < 0$. The process is spontaneous.

Let’s match these signs to the options:

1. $\Delta G_{\text{sys}}$: Negative ($-$) — spontaneous mixing.
2. $\Delta S_{\text{sys}}$: Positive ($+$) — increased disorder.
3. $\Delta S_{\text{surr}}$: Zero ($0$) — no heat exchange with surroundings.
4. $\Delta H$: Zero ($0$) — no enthalpy change for ideal mixing.

Only option (C) shows exactly this pattern: $\Delta G$ (–), $\Delta S_{\text{sys}}$ (+), $\Delta S_{\text{surr}}$ (0), $\Delta H$ (0).

An ideal solution obeys Raoult's law at all concentrations and temperatures. The key condition is that the solute-solvent interactions (A–B) must be equal in strength to the pure component interactions (A–A and B–B). When this holds, there is no volume change or enthalpy change on mixing — the solution is "ideal."

Let’s examine each pair.

1. Option (A): $\mathrm{CCl}_4 + \mathrm{C}_7\mathrm{H}_8$ (carbon tetrachloride + toluene)

   $\mathrm{CCl}_4$ is a non-polar, symmetrical molecule. Toluene ($\mathrm{C}_7\mathrm{H}_8$) is also non-polar but has a slightly polarizable aromatic ring. While both are non-polar, their molecular shapes and sizes differ enough that the intermolecular forces are not perfectly matched. In fact, $\mathrm{CCl}_4$ and toluene show slight positive deviation from Raoult’s law — not ideal.

2. Option (B): $\mathrm{CHCl}_3 + \mathrm{C}_6\mathrm{H}_6$ (chloroform + benzene)

   Chloroform has a polar C–H bond and can form weak hydrogen bonds with benzene’s π-electron cloud. This creates stronger A–B interactions than the pure A–A or B–B interactions. The result is a negative deviation from Raoult’s law — definitely not ideal.

3. Option (C): $\mathrm{H}_2\mathrm{O} + \mathrm{CH}_3\mathrm{OH}$ (water + methanol)

   Both are strongly hydrogen-bonded. However, water’s hydrogen-bond network is more structured than methanol’s. When mixed, the interactions are not identical — there is a significant enthalpy change and volume contraction. This mixture shows positive deviation and is far from ideal.

4. Option (D): $n-\mathrm{C}_6\mathrm{H}_{14} + n-\mathrm{C}_7\mathrm{H}_{16}$ ($n$-hexane + $n$-heptane)

   Both are straight-chain alkanes — non-polar, with nearly identical intermolecular forces (London dispersion forces). Their molecular sizes are similar, and the C–H bonds are alike. When mixed, the A–B interactions are essentially the same as A–A and B–B. This mixture obeys Raoult’s law very closely and is a textbook example of an ideal solution.