Chemistry Q56 – JEE-MAIN 2026

When

0.25

moles of a non-volatile, non-ionizable solute was dissolved in

1

mole of a solvent the vapor pressure of solution was

x

\% of vapor pressure of pure solvent. What is

x

\%?

Clue 1 of 3: Colligative Properties

Raoult's Law describes the vapor pressure of a solution containing a non-volatile solute.

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Ninja Strategy•Direct Raoult's Law Application

For a non-volatile solute, the vapor pressure of the solution is directly proportional to the mole fraction of the solvent, so calculate the solvent's mole fraction and convert it to a percentage.

🧠 Full Solution Path

Step 1: Calculate total moles and mole fraction of solvent✦ Active

The total number of moles in the solution is the sum of moles of solute and solvent.

n_{t o t a l} = n_{s o l u t e} + n_{s o l v e n t} = 0.25 mol + 1 mol = 1.25 mol

The mole fraction of the solvent ( $X_{s o l v e n t}$ ) is calculated as:

X_{s o l v e n t} = \frac{n_{s o l v e n t}}{n_{t o t a l}} = \frac{1 mol}{1.25 mol} = 0.8

Step 2: Apply Raoult's Law○ Expand

According to Raoult's Law for a non-volatile solute, the vapor pressure of the solution ( $P_{s o l u t i o n}$ ) is equal to the mole fraction of the solvent multiplied by the vapor pressure of the pure solvent ( $P_{s o l v e n t}^{0}$ ).

P_{s o l u t i o n} = X_{s o l v e n t} \cdot P_{s o l v e n t}^{0} = 0.8 \cdot P_{s o l v e n t}^{0}

Step 3: Determine the percentage○ Expand

The vapor pressure of the solution is $0.8$ times the vapor pressure of the pure solvent. To express this as a percentage, multiply by $100$ .

x \% = \frac{P_{s o l u t i o n}}{P_{s o l v e n t}^{0}} \times 100 % = \frac{0.8 \cdot P_{s o l v e n t}^{0}}{P_{s o l v e n t}^{0}} \times 100 % = 0.8 \times 100 % = 80 %

Therefore, $x = 80$ .

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Chemistry Question 56 – JEE-MAIN 2026

Clue 1 of 3: Colligative Properties

🧠 Full Solution Path