$11 g$ of an ideal gas contains only atoms of $C$ and $O$ occupies $5.6 L$ at $N T P$ . Identify the gas.

C H_{4}

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N_{2}

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O_{2}

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C O_{2}

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Solution

The correct option is D $C O_{2}$
$N T P$ is commonly used as a standard condition for testing and documentation of fan capacities: $N T P$ - normal temperature and pressure - is defined as air at $20^{o} C$ ( $293.15 K$ , $68^{o} F$ ) and $1 a t m$ ( $101.325 k P a$ ) density $1.204 k g / m^{3}$ ( $0.075$ pounds per cubic foot)

At $N T P$ , $P = 101325 P a$ and $V = 5.6 L$ given,so
Gas law is $P V = n R T$
$\Rightarrow \frac{P V}{R T} = \frac{101325 \times 5.6 \times 0.001}{8.314 \times 293.15} = 0.23 \sim 0.25$

$\frac{m}{M} = 0.23$
$M = \frac{m}{0.23} = 11 / 0.23 = 45 g$
$C O_{2}$ has mol wt. 44.
So D is correct answer.

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