# ques no 13 and 14 with explaination

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 *10

^{23})

22.4 L - has 6.02 *10

^{23}molecules the,

15 L - has 6.02*10

^{23}*15/22.4 = 4.03*10

^{23 }molecules

again 22.4 L - has N

_{A}molecules

5 L - 6.02*10

^{23}*5/22.4 = 1.34 *10

^{23}molecules

molecular mass of hydrogen gas is 2 g

2g - has N

_{A}molecules

0.5 g = N

_{A}*0.5/2 = 1.5*10

^{23}molecules

molecular mass of oxygen gas is 32 g

10g = N

_{A}*10/32 = 1.88*10

^{23}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*10

^{23}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*10

^{23}atoms.

Now if 1 mole - has 4*N

_{A}atoms

then 0.1 mole will have 0.4*N

_{A}atoms - 2.4*10

^{23}atoms

That is option (3)

Hope this helped ! :))

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