Solubility of N2 in water at 25°C is 6.8 × 10⁻⁴ mol/L at 1 atm partial pressure. By Henry's law, at 5 atm partial pressure of N2:
A6.8 × 10⁻⁴
B6.8 × 10⁻⁵
C3.4 × 10⁻³
D6.8 × 10⁻²
Answer & Solution
Correct answer: C. 3.4 × 10⁻³
By Henry's law: C ∝ P. At 5 atm: C = 5 × 6.8 × 10⁻⁴ = 3.4 × 10⁻³ mol/L.
Keep practising Unit 3B1: Solutions and Concentrations
Browse all Unit 3B1: Solutions and Concentrations questions for JEE Advanced, or explore more Chemistry topics.
Related questions
Hydrogenation of vegetable oils to make margarine uses catalyst:Adsorption isotherm:Freezing-point depression and elevation are colligative properties: depend on:For a 1 m glucose aqueous solution (i = 1), boiling point:For salt CaCl₂ (assume full dissociation), i =For 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵), van't Hoff factor i is:For solution of urea (M_w = 60 g/mol) with 6 g in 100 g water, molality:Osmotic pressure of 0.01 M sucrose solution at 27°C (R = 0.0821 L atm/mol K):