Question

How many grams are in 44.8 liter of oxygen gas
at STP?

Answer 1

Answer:

Mole of o2=44.8/22.4=2

1mole atom =6.022*10^23 atom

Then

no of atom in 2 mole o2=======2*6.022*10^23=12.044 *10^23

Answer 2

Answer:

Pressure= 2 atm

temperature= 273K

volume= 44.8cm

3

=44.* litre

so,By PV=nRT

n=

RT

PV

mole=

0.0821×273

2×44.8

so, no of moles= 3.998=4

hence, 4×6.023×10

23

=24.08×10

23

Explanation:

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