Calculate the mass of 120 cc of nitrogen at NTP . How many numbere of mopecules are present in it?
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1cm3 = 1/1000000 m3
120 cm 3 = 120/1000000 m3
1 m3 = 1000 L
120/1000000 m3 = 120/1000 L
Moles = volume/22.4
Moles = 120/22400
Moles = 0.0053 mols
Number of atoms in 1 mole is 6.023 * 10^23
Number of atoms in 0.0053 moles ==
3.22 * 10^21
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