Chemistry Q73 – JEE-MAIN 2025

When 1 g each of compounds AB and AB

_{2}

are dissolved in 15 g of water separately, they increased the boiling point of water by 2.7 K and 1.5 K respectively. The atomic mass of A (in amu) is _______

\times 10^{- 1}

(Nearest integer) (Given : Molal boiling point elevation constant is 0.5 K kg mol

^{- 1}

)

Enter Calculated Numerical Value:

Clue 1 of 3: Colligative Properties

The problem involves the elevation of boiling point, a colligative property that depends on the number of solute particles, not their identity.

🧠 Full Solution Path

Step 1: Calculate Molar Masses for AB and AB

_{2}

✦ Active

Given the boiling point elevation values and the molal boiling point elevation constant ( $K_{b} = 0.5$ K kg mol $^{- 1}$ ), we use the formula $Δ T_{b} = i \cdot K_{b} \cdot m$ . The mass of solvent (water) is $15$ g $= 0.015$ kg. Since assuming dissociation ( $i = 2$ for AB and $i = 3$ for AB $_{2}$ ) leads to a negative atomic mass, we assume the compounds behave as non-electrolytes, so the van't Hoff factor $i = 1$ for both.

For AB:

Δ T_{b} = 2.7 K

2.7 = 1 \times 0.5 \times \frac{1 g}{(M_{A} + M_{B}) \times 0.015 kg}

M_{A} + M_{B} = \frac{0.5}{2.7 \times 0.015} = \frac{0.5}{0.0405} \approx 12.3457 g/mol \quad (Equation 1)

For AB

_{2}

Δ T_{b} = 1.5 K

1.5 = 1 \times 0.5 \times \frac{1 g}{(M_{A} + 2 M_{B}) \times 0.015 kg}

M_{A} + 2 M_{B} = \frac{0.5}{1.5 \times 0.015} = \frac{0.5}{0.0225} \approx 22.2222 g/mol \quad (Equation 2)

Step 2: Solve the System of Equations for Atomic Masses○ Expand

We have a system of two linear equations for $M_{A}$ and $M_{B}$ :

M_{A} + M_{B} = 12.3457

M_{A} + 2 M_{B} = 22.2222

Subtract Equation 1 from Equation 2:

(M_{A} + 2 M_{B}) - (M_{A} + M_{B}) = 22.2222 - 12.3457

M_{B} = 9.8765 amu

Substitute

M_{B}

back into Equation 1:

M_{A} + 9.8765 = 12.3457

M_{A} = 12.3457 - 9.8765 = 2.4692 amu

💡 Teacher's Secret Hint

Ensure to carry enough decimal places during intermediate calculations to maintain precision for the final rounding.

Step 3: Determine the Final Answer○ Expand

The atomic mass of A is $2.4692$ amu. The question asks for the value in the blank, which is $M_{A} \times 10^{1}$ (Nearest integer).

\text{Value} = M_A \times 10^1 = 2.4692 \times 10 = 24.692

Rounding to the nearest integer, the answer is $25$ .

💡 Teacher's Secret Hint

Pay close attention to the format requested for the final answer, especially the multiplication factor and rounding instructions.

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Chemistry Question 73 – JEE-MAIN 2025

Clue 1 of 3: Colligative Properties

🧠 Full Solution Path