Solutions — NEET UG practice

1. Recall that for an ideal solution, mixing two liquids does not change the enthalpy or volume of the system:

$$\Delta_{mix}H = 0 \quad \text{and} \quad \Delta_{mix}V = 0$$

This matches exactly what Assertion A states, so A is correct.

2. Now check the reasoning given in Reason R, which claims that no interactions occur between P and Q at all.

In reality, an ideal solution does have P–Q interactions; the key condition is that the energy required to break the P–P and Q–Q interactions is exactly equal to the energy released when new P–Q interactions form.

Since the interactions are equal in strength (not absent), $\Delta_{mix}H$ comes out to be zero — not because there are no interactions.

3. Because R gives a false reason (interactions do exist; they are just balanced), R is incorrect.

4. Since A is true and R is false, the correct choice is that A is correct but R is not correct.

Hence, the answer is option D.

1. Check statement (A) using the molality formula, defined as moles of solute divided by mass of solvent in kilograms.

Given 2.5 g of ethanoic acid with molar mass $60 \, g \, mol^{-1}$, dissolved in 75 g (0.075 kg) of benzene, the moles of solute are

$$n = \frac{2.5}{60}$$

Therefore, the molality is

$$m = \frac{2.5/60}{0.075} = 0.556 \, mol \, kg^{-1}$$

So statement (A) is correct.

2. Check statement (B) using the molarity formula, which is moles of solute divided by volume of solution in litres.

Given 5 g of NaOH with molar mass $40 \, g \, mol^{-1}$ in 450 mL (0.45 L) of solution, the moles are

$$n = \frac{5}{40}$$

Hence, the molarity is

$$M = \frac{5/40}{0.45} = 0.278 \, mol \, L^{-1}$$

So statement (B) is correct.

3. Check statement (C) using Henry's law, which relates the solubility of a gas in a liquid to temperature through the Henry's law constant $K_H$.

Since $K_H$ increases with temperature while solubility is inversely related to $K_H$, gas solubility (like dissolved oxygen) is higher in cold water.

This means aquatic species indeed find cold water more comfortable because more oxygen is dissolved in it, so statement (C) is correct.

4. Check statement (D) using Henry's law in the form

$$P = K_H \, x$$

where $P$ is the partial pressure of the gas and $x$ is its mole fraction (solubility) in the liquid.

Since $P$ and $x$ are directly proportional, solubility actually increases with an increase in pressure, not a decrease.

So statement (D) is incorrect.

5. Check statement (E) using the definition of mole fraction, which for component B in a binary mixture of A and B is

$$x_B = \frac{n_B}{n_A + n_B}$$

The statement given in the question incorrectly places $n_A$ in the numerator instead of $n_B$, so statement (E) is incorrect.

6. Since only (A), (B) and (C) are correct, the matching option is A, B and C only.

Hence, the answer is option A.

1. Recall that a solution shows negative deviation from Raoult's law when the interactions between two different types of molecules (unlike molecules) are stronger than the interactions between molecules of the same kind.

2. In a chloroform–acetone mixture, the hydrogen atom of chloroform (made more acidic by the three electronegative chlorine atoms) forms a hydrogen bond with the lone pair of electrons on the oxygen atom of acetone:

$$CHCl_3 \cdots O=C(CH_3)_2$$

This new chloroform–acetone hydrogen bond is stronger than the original chloroform–chloroform and acetone–acetone interactions.

3. Because the molecules are held together more strongly than before, their escaping tendency decreases, so fewer molecules escape into the vapour phase.

4. A lower escaping tendency means the vapour pressure of the mixture is lower than the value predicted by Raoult's law, which is exactly what negative deviation means, and it also explains the corresponding increase in boiling point of the mixture.

5. This confirms that the negative deviation arises from hydrogen bond formation between acetone and chloroform, not from repulsive forces or increased escaping tendency.

Hence, the answer is option B.