g = 1.964×10−3 g/cc (1 dm = 10 cm) Temperature, T = 273 K Pressure, P = 76 cm Hg 76 cm = 760 mm = 1 atm (Since 1 cm = 10 mm and 760 mm = 1 atm) Gas constant, R = 0.0821 L atm‌K−1‌mol−1 = 0.0821 × 103 cc atm‌K−1‌m‌o‌l−1 (Since 1L =1000 cc) = 82.1 cc atm‌K−1‌mol−1 Now, ideal gas equation is: PV = nRT where, n = Number of moles Let given mass of the gas be 'W' g and molar mass be 'M' g. PV =
W
M
RT (Since, no. of moles = Given‌mass∕Molar‌mass) M =
W×RT
V×P
= d.
RT
P
(density = W∕V) Substituting all the value in the above equation, we get M =
1.964×0−3×82.1×273
1
= 44 g/mol Thus, the gas is CO2, as the molar mass of CO2 is 44 g/mol. Hence, option 'C' is correct.