Density of dry air (only $N_{2}$ & $O_{2}$ ) is $1.146 g l i t r e^{- 1}$ at $740 m m$ and $300 K$ . Calculate composition of air by weight assuming ideal nature.

Open in App

Solution

Let us assume the volume to be $1$ litre
$∴ 1.146 = \frac{m}{1}$
$m = 1.146 g$

Now, $P V = n R T$

$740 m m = 0.97$ atm
$∴ 0.97 \times 1 = \frac{m_{m i x}}{M_{m i x}} \times 0.0821 \times 300$
$∴ M_{m i x} = \frac{1.146 \times 0.0821 \times 300}{0.97}$
$∴ M_{m i x} = 29.1$

But $M_{m i x} = X_{O_{2}} \times M_{O_{2}} + X N_{2} \times M_{N_{2}}$
where $X_{O_{2}}$ & $X_{N_{2}}$ are the mole fractions of $O_{2}$ and $N_{2}$ respectively.

$∴ 29.1 = 32 \times X_{O_{2}} + 28 \times X_{N_{2}} ⟶ (i)$
and $X_{O_{2}} + X_{N_{2}} = 1 ⟶ (i i)$

Solving (i) & (2), we get,
$X_{O_{2}} = 0.275$ and $X_{N_{2}} = 0.725$

Weight % of $O_{2} = \frac{0.275 \times 32}{(0.275 \times 32) + (0.725 \times 28)} \times 100 = 30.24$ %

$∴$ Weight % of $N_{2} = 100 - 30.24 = 69.76$ %
$∴$ Composition of air is $O_{2} = 30.24$ %
$N_{2} = 69.76$ % {by weight}

Suggest Corrections

Similar questions

density of dry air containing only N2 and O2 is 1.15 g/l at 740mm of hg and 300 k what is the % composition of N2 by mass in the air

N_{2}

O_{2}

1.15 g / L

740

300

N_{2}

22 % O_{2}

78 % N_{2}

80

N_{2}

20

O_{2}

293 K

76 c m H g

22 % O_{2}

78 % N_{2}

Join BYJU'S Learning Program