20. if average molecular weight of air is 29, then assuming N2 and O2 gases are there which option are correct regarding composition of air : 1] 75% by mass ofN2 2] 75% by moles of N2 3] 72.41% by mass of N2

Open in App

Solution

$M_{a v g} = \frac{n_{N_{2}} \times M_{N_{2}} + (100 - n_{N_{2}}) \times M_{O_{2}}}{100}$
$29 = \frac{n \times 28 + (100 - n) \times 32}{100}$
$2900 = 28 n + 3200 - 32 n = 4 n = 300$
$M o l e p e r c e n t a g e o f N_{2} = 75 %$
$M o l e p e r c e n t a g e o f O_{2} = 25 %$
$\frac{n_{N_{2}}}{n_{O_{2}}} = \frac{3}{1} = y_{N_{2}} = 0.75, y_{O_{2}} = 0.25$
$W_{N_{2}} = 0.75 \times M_{N_{2}} = 0.75 \times 28 = 21 g r a m s$
$W_{O_{2}} = 0.25 \times M_{O_{2}} = 0.25 \times 32 = 8 g r a m s$
$\frac{% w}{w} o f N_{2} = \frac{21}{21 + 8} \times 100 = \frac{72.41 % w}{w} o f N_{2}$
$\frac{% w}{w} o f O_{2} = \frac{8}{21 + 8} \times 100 = \frac{27.59 % w}{w} o f O_{2}$

Suggest Corrections

Similar questions

N_{2}

O_{2}

N_{2}

N_{2}

N_{2}

N_{2}

O_{2}

1.146 g l i t r e^{- 1}

740 m m

300 K

80

N_{2}

20

O_{2}

293 K

76 c m H g

22 % O_{2}

78 % N_{2}

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

Join BYJU'S Learning Program