# ques no 13 and 14 with explaination

Ok so 13. first,
Now maximum no of molecules present in which of the 4 is the question.
So any molecular mass of gas at STP occupies a volume of 22.4 litres and has avogadro number molecules ( 6.023 *1023)

22.4 L - has 6.02 *1023 molecules the,
15 L - has  6.02*1023*15/22.4 = 4.03*1023 molecules

again 22.4 L - has NA molecules
5 L - 6.02*1023*5/22.4 = 1.34 *1023 molecules

molecular mass of hydrogen gas is 2 g
2g - has NA molecules
0.5 g = NA*0.5/2 = 1.5*1023 molecules

molecular mass of oxygen gas is 32 g
10g = NA*10/32 = 1.88*1023 molecules.

Hence option (1) has the highest no of molecules.

Now coming to ques 14.
Generally we assume 1 mole of a gas to have 1*6.02*1023 atoms - since it is assumed to be 'mono' atomic.
But here it is mentioned for the gas to be tetratomic  - that it within 1 mole there are already existing 4 atoms bound together - hence it would have 4*6.02*1023 atoms.

Now if 1 mole - has 4*NA atoms
then 0.1 mole will have 0.4*NA atoms - 2.4*1023 atoms

That is option (3)

Hope this helped ! :))

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Hope it helped:)

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