Solutions Previous Year Questions — NEET Chemistry

2 past-year Solutions questions from NEET (Chemistry).

Q76 (2024)

Identify the correct statements: (A) The molality of 2.5 g of ethanoic acid (Molar mass: $60 \text{ g mol}^{-1}$) in 75 g of benzene solution is 0.556 m. (B) The molarity of a solution containing 5 g of NaOH (molar mass: $40 \text{ g mol}^{-1}$) in 450 mL of solution is 0.278 M at 298 K. (C) Aquatic species are more comfortable in cold water. (D) The solubility of gas increases with decrease in pressure. (E) For a binary mixture of A and B, the number of moles of A and B are $n_A$ and $n_B$ respectively. The mole fraction of B will be $x_B = \frac{n_A}{n_A \times n_B}$. Choose the correct answer from the options given below:
  1. (1) A and C only
  2. (2) A, B and C only
  3. (3) A, D and E only
  4. (4) A and B only
### Related Formula $$m = \frac{w_B}{M_B} \times \frac{1000}{w_A(g)}, \quad M = \frac{w_B}{M_B} \times \frac{1000}{V(mL)}$$ ### Core Logic (A) $m = \frac{2.5}{60} \times \frac{1000}{75} = 0.556 \text{ m}$ (Correct). (B) $M = \frac{5}{40} \times \frac{1000}{450} = 0.278 \text{ M}$ (Correct). (C) Lower temp $\implies$ higher gas solubility $\implies$ more $O_2$ in cold water (Correct). (D) Gas solubility increases with increase in pressure (Incorrect). (E) $x_B = \frac{n_B}{n_A + n_B}$ (Incorrect). ### Step 1: Conclusion Statements A, B and C are correct (Option 2). ### Pattern Recognition Verify numerical formulas for molality, molarity, and Henry's law temperature dependence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions

Q77 (2024)

Mixture of chloroform and acetone forms a solution with negative deviation from Raoult's law due to:
  1. (1) Formation of hydrogen bonding between acetone and chloroform
  2. (2) Increase in escaping tendency of molecules of each component
  3. (3) Stronger intermolecular forces between chloroform molecules than those between chloroform and acetone molecules
  4. (4) Repulsive forces
### Related Formula $$CH_3-C(=O)-CH_3 \cdots H-CCl_3 \quad (\text{Intermolecular Hydrogen Bonding})$$ ### Core Logic Acetone and chloroform form intermolecular hydrogen bonding between oxygen of acetone and acidic hydrogen of chloroform. This increases $A-B$ interactions, causing negative deviation from Raoult's law. ### Step 1: Conclusion Reason is formation of hydrogen bonding (Option 1). ### Pattern Recognition Chloroform + Acetone $\rightarrow$ Hydrogen bonding $\rightarrow$ Stronger solute-solvent attractive forces $\rightarrow$ Negative deviation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Solutions
Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

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