Molality of 3M NaCl Solution – JEE Main Previous Year Question

JEE Main Previous Year Question

Q. Molality (m) of 3 M aqueous solution of NaCl is :

(Given : Density of solution = 1.25 g mL⁻¹,
Molar mass in g mol⁻¹ : Na = 23, Cl = 35.5)

A. 2.90 m

B. 3.85 m

C. 1.90 m

D. 2.79 m

Step 1: Assume volume of solution

Assume volume of NaCl solution = 1 L

Step 2: Calculate mass of solution

Density = 1.25 g mL⁻¹
Mass of 1 L solution = 1.25 × 1000 = 1250 g

Step 3: Calculate moles and mass of NaCl

Molarity = 3 M
Moles of NaCl = 3 moles
Molar mass of NaCl = 23 + 35.5 = 58.5 g mol⁻¹
Mass of NaCl = 3 × 58.5 = 175.5 g

Step 4: Calculate mass of solvent

Mass of solvent = Mass of solution − Mass of solute
= 1250 − 175.5 = 1074.5 g = 1.0745 kg

Step 5: Calculate molality

Molality (m) = moles of solute / kg of solvent
= 3 / 1.0745 = 2.79 m

Correct Answer: 2.79 m