The total number of moles present in a gaseous mixture containing $8 g$ of oxygen, $11.2 l i t r e s$ of $C O_{2}$ and $3 \times 10^{22}$ molecules of $N_{2}$ at STP is

1.25

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1.05

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0.80

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0.35

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Solution

The correct option is C
0.80

$n = \frac{W_{t}}{G . M . W_{t}} = \frac{G i v e n V o l u m e}{22.4 l i t} = \frac{G i v e n n o . o f m o l e c u l e s}{6.023 \times 10^{23}}$

$n = \frac{1}{4} + \frac{1}{2} + \frac{1}{20} = 0.8$

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The total number of moles present in a gaseous mixture containing 8 g of oxygen, 11.2 litres of CO2 and 3 × 1022 molecules of N2 at STP is

The correct option is C 0.80 n = WtG.M.Wt = Given Volume22.4lit = Given no. of molecules6.023 × 1023 n = 14 + 12 + 120 = 0.8

The total number of moles present in a gaseous mixture containing $8 g$ of oxygen, $11.2 l i t r e s$ of $C O_{2}$ and $3 \times 10^{22}$ molecules of $N_{2}$ at STP is

The correct option is C
0.80

$n = \frac{W_{t}}{G . M . W_{t}} = \frac{G i v e n V o l u m e}{22.4 l i t} = \frac{G i v e n n o . o f m o l e c u l e s}{6.023 \times 10^{23}}$

$n = \frac{1}{4} + \frac{1}{2} + \frac{1}{20} = 0.8$