Question

Calculate the no. of atoms present in 11.5 litres of H2 at N.T.P. give solution

Answer 1

No. Of mole=given volume in L/22.4L
=11.5/22.4
= 0.5
No of atoms = moles × avogadro number
=0.5×6.022 x 10^23
=3.011 x 10^23

Answer 2

Answer:

We know that, at standard temperature and pressure, 1 mole of any gas at STP contains 22.4 L of it.

1 mole of Mercury(Hg) at STP = 22.4L

Therefore 11.5 liters of hydrogen(H2) is (1/22.4)*11.5 moles  = 0.51 moles of hydrogen gas.

We know that in 1 mole there is 6.022 × 10^23 molecules.

Hence,  for  0.51 moles of Hydrogen =0.51*[6.022 × 10^23] molecules of Hydrogen gas.

As there are 1 atoms in 2 molecule of Hydrogen(H2).  

Thus no. of atoms of mercury =2x3.071x 10 ^23 atoms=6.14 x 1023 atoms of H2.