ques no 13 and 14 with explaination

ques no 13 and 14 with explaination :ontaining 3 g nolar mass of g isotopic 2.0%. The e naturally olecules at OOC (1) 7 g N20 (3) 16 g N02 V2fäö g H2 (4) 16 g S02 20. The number of molecules in 4.25 g of NH 3 approximately (l) 4 x 1023 ) 1 x 1023 0-45 x 1023 (4) 6 x 1023 The maximum number of molecules is present i (1) 15 L of H2 gas at STP (2) 5 L of N2 gas at STP (3) 0.5 g of H2 gas 4) 10 g of 02 gas The number of atoms in 0.1 mol of a tetraatom• gas is (NA = 6.02 x 1023 mol-l) (1) 2.4 x 1022 (3) 2.4 x 1023 (2) 6.026 x 1022 (4) 3.600 x 1023

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 N_A molecules
          5 L - 6.02*10²³*5/22.4 = 1.34 *10²³ molecules

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

molecular mass of oxygen gas is 32 g
10g = N_A*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*N_A atoms
then 0.1 mole will have 0.4*N_A atoms - 2.4*10²³ atoms 

That is option (3)

Hope this helped ! :))