$(a)\;1.5124g/l\qquad(b)\;151.24g/l\qquad(c)\;2.3g/l\qquad(d)\;2.094g/l$

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P = $\large\frac{800 \text{ mm Hg}}{760}=\large\frac{800}{760}atm$

T = 273+100 = 300K

Let the density be d for $CO_2$

From Ideal gas equation, for $CO_2$ PV = $\large\frac{w}{m}RT$

P = $\large\frac {d}{m} RT$

(Since d = $\large\frac{w}{v}$)

$\large\frac{800}{760} = \large\frac{d}{44}\times0.0821\times373$

$\therefore d = 1.5124g/l$

Hence answer is (a)

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