The volume occupied by $8.8 g C O_{2}$ at $31^{o} C$ and 1 atm pressure is

5 ml

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0.2 L

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5 L

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2 L

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Solution

The correct option is C 5 L
Here: Given mass of $C O_{2} = 8.8 gram$
Molar mass of $C O_{2} = 44 gram$
Now, number of moles of $C O_{2} = \frac{8.8}{44} = 0.2 moles$
$n = 2$
Using $P V = n R T$
$V = \frac{n R T}{P} =$ volume occupied.
$n = 0.2, R = \frac{0.0821 L - a t m}{K m o l}$
$T = 31^{o} C = 31 + 273 = 304 K$
$P = 1 atm$
$Volume = \frac{0.2 m o l \times 0.0821 \times L - a t m \times 304 K}{1 a t m \times K m o l}$
$= 0.2 \times 0.0821 \times 304 L$
$= 4.99$
Approx. $5$ Liters

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