Chemistry Q75 – JEE-MAIN 2026

A non-volatile, non-electrolyte solid solute when dissolved in

40 g

of a solvent, the vapour pressure of the solvent decreased from

760 mm Hg

750 mm Hg

. If the same solution boils at

320 K

, then the number of moles of the solvent present in the solution is _______. (Nearest integer) [Given: boiling point of the pure solvent =

319.5 K

K_{b}

of the solvent =

0.3 {K kg mol}^{- 1}

]

Enter Calculated Numerical Value:

Clue 1 of 3: Colligative Properties

The problem involves two colligative properties: relative lowering of vapor pressure and elevation in boiling point. Both depend on the number of solute particles.

🧠 Full Solution Path

Step 1: Calculate moles of solute (

n_{B}

) using elevation in boiling point✦ Active

The elevation in boiling point is calculated as the difference between the solution's boiling point and the pure solvent's boiling point:

Δ T_{b} = T_{b} - T_{b}^{0} = 320 K - 319.5 K = 0.5 K

Using the formula for elevation in boiling point, $Δ T_{b} = K_{b} \cdot m$ , we can find the molality ( $m$ ) of the solution:

m = \frac{Δ T_{b}}{K_{b}} = \frac{0.5 K}{0.3 {K kg mol}^{- 1}} = \frac{5}{3} {mol kg}^{- 1}

Now, calculate the moles of solute ( $n_{B}$ ) using the molality and the mass of the solvent ( $W_{A} = 40 g = 0.040 kg$ ):

n_{B} = m \times W_{A} = \frac{5}{3} {mol kg}^{- 1} \times 0.040 kg = \frac{0.2}{3} = \frac{1}{15} mol

Step 2: Calculate moles of solvent (

n_{A}

) using Raoult's Law○ Expand

Apply Raoult's Law for relative lowering of vapor pressure, which states that the relative lowering of vapor pressure is equal to the mole fraction of the solute ( $x_{B}$ ):

\frac{P_{A}^{0} - P_{A}}{P_{A}^{0}} = x_{B} = \frac{n_{B}}{n_{A} + n_{B}}

Substitute the given vapor pressures ( $P_{A}^{0} = 760 mm Hg$ , $P_{A} = 750 mm Hg$ ):

\frac{760 mm Hg - 750 mm Hg}{760 mm Hg} = \frac{10}{760} = \frac{1}{76}

Now, equate this to the mole fraction of the solute:

\frac{1}{76} = \frac{n_{B}}{n_{A} + n_{B}}

Rearrange the equation to solve for $n_{A}$ :

76 n_{B} = n_{A} + n_{B} ⟹ 75 n_{B} = n_{A}

Substitute the value of $n_{B}$ calculated in Step 1:

n_{A} = 75 \times \frac{1}{15} mol = 5 mol

Step 3: Final Answer○ Expand

The number of moles of the solvent present in the solution is $5 mol$ . Rounding to the nearest integer, the answer is $5$ .

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Chemistry Question 75 – JEE-MAIN 2026

Clue 1 of 3: Colligative Properties

🧠 Full Solution Path