The molecular weight of the gas is 9.9 g/mol, matching option d. This CSIR NET-style question tests ideal gas law application for gases occupying equal volumes at same pressure but different temperatures and masses.​

Problem Setup

4.8 g of unknown gas at 27°C (300 K) occupies the same volume as 1 g of hydrogen (molecular weight 2 g/mol) at 17°C (290 K), under identical pressure. Since pressure and volume are constant, the ideal gas law PV = nRT simplifies to n ∝ m/MT, where n is moles, m is mass, M is molecular weight, and T is absolute temperature. Equal volumes imply equal moles: m₁/(M × T₁) = m₂/(M_H₂ × T₂).​

Step-by-Step Solution

Convert temperatures to Kelvin: T₁ = 27 + 273 = 300 K, T₂ = 17 + 273 = 290 K. Rearrange for M: M = (m₁ × M_H₂ × T₁) / (m₂ × T₂) = (4.8 × 2 × 300) / (1 × 290) = 2880 / 290 ≈ 9.93 g/mol, closest to 9.9 g. This uses Avogadro’s law extension for non-STP conditions, as equal V and P mean equal n adjusted for T.​

Option Analysis

• a. 0.1 g: Far too low; implies unrealistically light gas, ignores mass ratio (4.8x heavier than H₂).​

• b. 0.48 g: Underestimates by not accounting full T₁/T₂ ratio (300/290 ≈ 1.034); wrong if T ignored.​

• c. 5.2 g: Possible if T₂ misread as higher or arithmetic error (e.g., 4.8 × 2 / 1.85), but incorrect.​

• d. 9.9 g: Correct; precise match after T adjustment, common in CSIR NET gas law problems.​

CSIR NET Tips

Practice ideal gas law variations like PV = (m/M)RT for molecular weight directly. Common pitfalls: forgetting Kelvin conversion or assuming same T. Relate to biochemistry contexts like gas diffusion in cells.​​