73. The number of atoms in 4.25 g of NH3 is approximately(c) 4x 102(a)

1.	73. The number of atoms in 4.25 g of NH3 is approximately(c) 4x 102(a) 1 1023(b) 2 x 10236 x 1023
Answer» 1 mole of Ammonia is 17gm. {NH3= N(14gm)+ H(1gm)+H(1gm)+H(1gm)} Now 1 mole of any substance at STP always has Avogadro's number of molecules i.e. 6.02310^23. So 1 mole or 17gm of Ammonia will have 6.02323 molecules of NH3. Here it's 4.25gm or 4.25/17= 0.25moles of Ammonia. Therefore 0.25 moles of Ammonia will have 0.256.02310^23 molecules. As 1 molecule has 4 atoms. (1 of Nitrogen and 3 hydrogen), We will have 4* 0.25* 6.02310^23 atoms in 0.25 moles of Ammonia. Answer is 4.25gms will have 6.02310^23 number of atoms in it.

73. The number of atoms in 4.25 g of NH3 is approximately(c) 4x 102(a) 1 1023(b) 2 x 10236 x 1023

Answer»

1 mole of Ammonia is 17gm.

{NH3= N(14gm)+ H(1gm)+H(1gm)+H(1gm)}

Now 1 mole of any substance at STP always has Avogadro's number of molecules i.e. 6.023*10^23.

So 1 mole or 17gm of Ammonia will have 6.023*23 molecules of NH3.

Here it's 4.25gm or 4.25/17= 0.25moles of Ammonia.

Therefore 0.25 moles of Ammonia will have 0.25*6.023*10^23 molecules.

As 1 molecule has 4 atoms. (1 of Nitrogen and 3 hydrogen),

We will have 4* 0.25* 6.023*10^23 atoms in 0.25 moles of Ammonia.

Answer is 4.25gms will have 6.023*10^23 number of atoms in it.