JEE Main 2026: Calculation of Atomic Masses of Elements P and Q using Elevation in Boiling Point
Q. Elements P and Q form two types of non-volatile, non-ionizable compounds PQ and PQ₂. When 1 g of PQ is dissolved in 50 g of solvent 'A', ΔTb was 1.176 K while when 1 g of PQ₂ is dissolved in 50 g of solvent 'A', ΔTb was 0.689 K. (Kb of 'A' = 5 K kg mol⁻¹). The molar masses of elements P and Q (in g mol⁻¹) respectively, are :
Correct Answer: A (25, 60)

Step-by-Step Derivation

Step 1: Formula for Molar Mass from Elevation in Boiling Point
The elevation in boiling point is given by: \( \Delta T_b = K_b \times m \)
Where \( m \) (molality) \( = \frac{w_2 \times 1000}{M_2 \times w_1} \)
Rearranging for Molar Mass (\( M_2 \)):
\( M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1} \)
Step 2: Calculate Molar Mass of Compound PQ
Given: \( w_2 = 1 \text{ g} \), \( w_1 = 50 \text{ g} \), \( K_b = 5 \), \( \Delta T_b = 1.176 \text{ K} \)
\( M_{PQ} = \frac{5 \times 1 \times 1000}{1.176 \times 50} \)
\( M_{PQ} = \frac{5000}{58.8} \approx 85.03 \text{ g/mol} \)
Let atomic mass of P be x and Q be y.
Equation 1: \( x + y = 85 \)
Step 3: Calculate Molar Mass of Compound PQ₂
Given: \( w_2 = 1 \text{ g} \), \( w_1 = 50 \text{ g} \), \( K_b = 5 \), \( \Delta T_b = 0.689 \text{ K} \)
\( M_{PQ_2} = \frac{5 \times 1 \times 1000}{0.689 \times 50} \)
\( M_{PQ_2} = \frac{5000}{34.45} \approx 145.13 \text{ g/mol} \)
Equation 2: \( x + 2y = 145.13 \)
Step 4: Solve the simultaneous equations
Subtract Eq 1 from Eq 2:
\( (x + 2y) - (x + y) = 145.13 - 85.03 \)
\( y = 60.1 \approx 60 \text{ g/mol} \)

Substitute \( y = 60 \) in Eq 1:
\( x + 60 = 85 \)
\( x = 25 \text{ g/mol} \)
Step 5: Final Conclusion
Atomic mass of P = 25 g/mol
Atomic mass of Q = 60 g/mol
Correct Option: A

Extensive Related Theory

1. Colligative Properties: An Overview
Colligative properties are those properties of solutions that depend solely on the number of solute particles (ions or molecules) present in a given amount of solvent and not on their chemical nature. There are four primary colligative properties studied in physical chemistry: - Relative lowering of vapor pressure. - Elevation in boiling point (\( \Delta T_b \)). - Depression in freezing point (\( \Delta T_f \)). - Osmotic pressure (\( \pi \)). In this problem, we specifically use the Elevation in Boiling Point to determine the molecular formula components.

2. Elevation in Boiling Point (\( \Delta T_b \))
The boiling point of a liquid is the temperature at which its vapor pressure becomes equal to the atmospheric pressure. When a non-volatile solute (like PQ or PQ₂) is added to a solvent, the vapor pressure of the solution decreases. To reach atmospheric pressure, the solution must be heated to a higher temperature than the pure solvent. This difference is known as the elevation in boiling point. The relation is: \( \Delta T_b = T_b - T_b^\circ \), where \( T_b \) is the boiling point of the solution and \( T_b^\circ \) is the boiling point of the pure solvent.

3. Ebullioscopic Constant (\( K_b \))
The proportionality constant \( K_b \) is known as the Molal Elevation Constant or the Ebullioscopic Constant. It is defined as the elevation in boiling point produced when 1 mole of solute is dissolved in 1 kg of solvent. The value of \( K_b \) depends only on the nature of the solvent and can be calculated using the formula: \[ K_b = \frac{R \cdot M_1 \cdot (T_b^\circ)^2}{1000 \cdot \Delta_{vap}H} \] Where \( R \) is the gas constant, \( M_1 \) is the molar mass of the solvent, and \( \Delta_{vap}H \) is the enthalpy of vaporization.

4. Non-Volatile and Non-Ionizable Nature
The question mentions that the compounds are "non-volatile" and "non-ionizable". - Non-volatile: Means the solute itself does not contribute to the vapor pressure. - Non-ionizable: This is a crucial hint! It means the van't Hoff factor (\( i \)) is 1. If the compound dissociated (like NaCl) or associated (like Acetic acid in benzene), the number of particles would change, and the formula would become \( \Delta T_b = i \cdot K_b \cdot m \). Since \( i=1 \), we can use the simplified formula safely.

5. Binary Compounds and Atomic Mass Determination
This specific problem type is a "puzzle" variant. Instead of asking for a single molar mass, the examiner provides two different compounds (\( PQ \) and \( PQ_2 \)) formed by the same elements. This requires setting up a system of linear equations. This approach is frequently used in JEE to test both your chemistry concepts and your algebraic speed. Key Assumptions: - The reaction to form PQ and PQ₂ goes to completion. - The solutions behave ideally (valid for dilute solutions).

6. Practical Tips for JEE Main 2026
- Unit Conversion: Ensure the mass of solvent is in kilograms if you are not using the "1000" factor in the numerator. - Rounding Off: In competitive exams, slight variations in decimals (like 145.1 vs 145) are common. Look for the closest integer match. - Reverse Verification: If time permits, plug the values 25 and 60 back into the equations to see if they yield the given \( \Delta T_b \).

Related JEE Main Questions

Given below are two statements : Statement I : The number of species among SF4, NH4+, [NiCl4]2-, XeF4, [PtCl4]2-, SeF4 and [Ni(CN)4]2-, that have tetrahedral geometry is 3 . Statement II : In the set [ NO2, BeH2, BF3, AlCl3 ], all the molecules have incomplete octet around central atom. In the light of the above statements, choose the correct answer from the options given below :
Use the following data : Substance | ΔfH°(500 K) / kJ mol⁻¹ | S°(500 K) / JK⁻¹ mol⁻¹ AB(g) | 32 | 222 A2(g) | 6 | 146 B2(g) | x | 280 One mole each of A2(g) and B2(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K . A2(g) + B2(g) ⇌ 2AB(g) The value of x (in kJ mol⁻¹) is _____. (Nearest integer) (Given : log K = 2.2, R = 8.3 J K⁻¹ mol⁻¹ )
Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mol⁻¹. If k1 and k2 are the rate constants of first and second reaction respectively at 300 K , then ln (k2/k1) will be _____. (nearest integer) [R = 8.3 J K⁻¹ mol⁻¹]
The pH and conductance of a weak acid (HX) was found to be 5 and 4 × 10⁻⁵ S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm² were at a distance of 15 cm apart. The value of the limiting molar conductivity is _____ Sm² mol⁻¹. (nearest integer) (Given : degree of dissociation of the weak acid (α) ≪ 1 )
An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br2 and KOH forms compound (R) having molecular formula C6H7N. Names of P, Q and R respectively are:

Related Covered Topics

colligative properties elevation in boiling point ebullioscopic constant molar mass determination van't hoff factor dilute solutions physical chemistry numericals jee mains 2026 preparation solvent properties ideal solutions molality calculation atomic mass calculation binary compounds chemistry solutions chapter competitive exam chemistry

About the Author

Jee Neet Experts – Senior academic contributors with a specialization in Physical Chemistry. We focus on providing high-fidelity solutions for the 2026 JEE Main shift papers, ensuring every step is derived from fundamental principles.

Home / Explore JEE & NEET Prep Material – neetjeerankers.com
Scroll to Top