UPSEE Solved Model Paper 7

Given: Density, d = 1.964 $\mathrm{g/dm}^{3}$ = $\frac{1.964}{10^{3}\,\mathrm{cc}}\,\mathrm{g}$ = $1.964\times 10^{-3}$ 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 $\mathrm{atm\,K^{-1}\,mol^{-1}}$ = 0.0821 × $10^{3}$ cc $\mathrm{atm\,K^{-1}\,mol^{-1}}$ (Since 1L =1000 cc) = 82.1 cc $\mathrm{atm\,K^{-1}\,mol^{-1}}$ 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 = $\frac{W}{M}RT$ (Since, no. of moles = $\frac{\text{Given mass}}{\text{Molar mass}}$) M = $\frac{W\times RT}{V\times P}$ = $\frac{d\cdot RT}{P}$ (density = $\frac{W}{V}$) Substituting all the value in the above equation, we get M = $\frac{1.964\times 0^{-3}\times 82.1\times 273}{1}$ = 44 g/mol Thus, the gas is $\mathrm{CO}_{2}$, as the molar mass of $\mathrm{CO}_{2}$ is 44 g/mol. Hence, option 'C' is correct.