Molarity from Mass Percentage – JEE Main PYQs | Concentration of Solutions

The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is ______ × 10−1 M. (Nearest integer)

Given: Density of aqueous solution of (X) is 1.25 g mL⁻¹
Molar mass of the acid is 70 g mol⁻¹

Step 1: Consider 100 g of the solution

70% (m/m) means 70 g of acid is present in 100 g of solution.

Step 2: Calculate volume of 100 g solution

Density = mass / volume
Volume = 100 / 1.25 = 80 mL = 0.08 L

Step 3: Calculate number of moles of acid

Moles of acid = 70 / 70 = 1 mole

Step 4: Calculate molarity of the solution

Molarity = moles / volume = 1 / 0.08 = 12.5 M

Step 5: Express in required format

12.5 M = 125 × 10−1 M

✅ Final Answer: 125
An organic compound weighing 500 mg produced 220 mg of CO₂ on complete combustion. The percentage composition of carbon in the compound is ______ %
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On complete combustion 1.0 g of an organic compound gave 1.46 g of CO₂ and 0.567 g of H₂O. The empirical formula mass of compound is ______ g
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Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.

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