2026 · Paper 2 · Chemistry

Question 16

The printed question, the official key, and each model’s JSON answer plus the reasoning it wrote.

ChemistryPaper 2Numerical

Q16

Official 0.16

JEE Advanced 2026 Paper 2 Chemistry question 16, page 1JEE Advanced 2026 Paper 2 Chemistry question 16, page 2

Model answers

Claude Fable 5.1

Anthropic

Correct2 / 2

0.16

Claude Fable 5.1 reasoning

5 molal solution of B in A: 5 mol B per 1000 g A → moles of A = 1000/50 = 20. Mole fraction of B in liquid: x_B = 5/25 = 0.2, x_A = 0.8. Total pressure: P = x_A P_A° + x_B P_B° → 100 = 0.8(105) + 0.2 P_B° → 100 = 84 + 0.2 P_B° → P_B° = 80 mm Hg. Partial pressure of B = 0.2 × 80 = 16 mm Hg. Mole fraction of B in vapour = 16/100 = 0.16.

GPT 6 Astra

OpenAI

Correct2 / 2

0.16

GPT 6 Astra reasoning

**Calculating fractions** I need to calculate the fraction involving nA20, nB5, xA.8, pA84, and pB16, and it seems like finalizing the output in a JSON format might be best. I want to verify there's no catch with the molality of 5. It's essential to ensure my calculations and formatting are correct for clarity and accuracy. I just want to make sure everything is precise before I wrap this up!

GPT 5.5

OpenAI

Correct2 / 2

0.16

GPT 5.5 reasoning

**Parsing solution details** I need to focus on solving this problem and producing the final JSON output. The task involves a 5 molal solution of volatile liquids A and B in a closed container, with a total vapor pressure of 100 mm Hg at 300 K. The vapor pressure of pure A is 105 mm Hg. In this case, I should find the mole fraction of B in the vapor phase, which requires careful consideration. A means 5 moles of B in 1 kg of A! **Calculating vapor pressures** I need to apply Raoult's law, where p_A equals x_A times the vapor pressure of pure A. I have a 5 molal solution of B in A, so I should calculate x_B from the molality. Given 1 kg of A, with a molar mass of 50 g/mol, I find n_A is 20 moles and n_B is 5 moles. This gives me x_B as 0.2 and x_A as 0.8, leading to a vapor pressure of A at 84 mmHg and B at 16 mmHg. However, since it's in a closed container, I wonder about possible changes in composition due to vaporization. **Determining vapor compositions** Given the total vapor pressure of the solution at equilibrium, I can calculate x_B using the formula for total pressure. I know p_total equals x_A times the vapor pressure of pure A plus x_B times the vapor pressure of B, which I need to find. Since I have x_B from the molality as 0.2, I can work backward. This means I have a solid starting point, but deriving PB* is still an unknown that I'll need to consider for the final answer. **Calculating equilibrium properties** I'm working through some calculations related to a 5 molal solution. It seems I should focus on the mole fraction of B in the vapor phase at equilibrium, using the formula P_B/P_total. I also need to ensure I deliver my final answer strictly as a JSON object, with an answer of 0.16. I should pay attention to numerical formatting since it’s asking for a mole fraction. Let's finalize this! {"mole_fraction": 0.16} **Analyzing vapor pressure** I'm examining the total vapor pressure, which is actually lower than that of pure A at 105 mmHg, even though B is volatile. Adding B decreases A's partial pressure by 21 mmHg to 84, and B contributes 16, resulting in a total depression of 5. This means B is less volatile, and its \( P_B^* \) is 80. The question mentions that A and B act as ideal gases in the vapor phase, which is essential for calculating the mole fraction via Dalton's law. I'm ready to finalize my answer!

Claude Sonnet 5

Anthropic

Correct2 / 2

0.16

Claude Sonnet 5 reasoning

I'm calculating the mole fractions from the molality: 20 mol solvent A and 5 mol solute B give x_A = 0.8 and x_B = 0.2, then using Raoult's law with P_A° = 105 mmHg and total pressure 100 mmHg to find P_A from x_A times P_A°. So P_B = 100 - 84 = 16 mmHg, giving mole fraction of B in vapor as 16/100 = 0.16. Using Raoult's law: x_A(liquid) = 20/25 = 0.8, x_B(liquid) = 5/25 = 0.2 P_A = x_A × P_A° = 0.8 × 105 = 84 mm Hg P_B = P_total − P_A = 100 − 84 = 16 mm Hg Mole fraction of B in vapor phase = P_B / P_total = 16/100 = 0.16 ```json