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Which of the following contains maximum number of atoms a)4.25g of NH3 b)8g of o2 c)2g of h2 d)4g of he

Which of the following contains maximum number of atoms a)4.25g of NH3 b)8g of o2 c)2g of h2 d)4g of he

Grade:11

5 Answers

Arun
25750 Points
6 years ago
Dear student
 
a) 4.25 gm of NH3
moles = 4.25/17 = 0.25
b) 8 gm of O2
moles = 8/32 = 0.25
c) 2gm of H2
moles = 2/4 = 0.5
d) 4 gm of He
moles = 4/4 = 1
Hence maximum number of atoms will be in option D.
 
Regards
Arun (askIITians forum expert)
praveen kumar
144 Points
6 years ago
3.2/2=1 mole=6.023 *10^23 molecules=2*6.023*10^23 atoms=12.046*10^23 atoms.
2g of h2 cantains 2 moles of hydrogen atoms so the answer is c. if you want to check for all do as like me. thank you.
Prajapati S
13 Points
5 years ago
Dear Students
  
(1) No. of atoms in 4.25 g NH3 = Moles * NA* atomicity
= 4.25/17  * NA * 4
= 1 NA
(2) No. of atoms in 8 g O2  = Moles *NA*atomicity
= 8/32 * NA *2
= 0.5 NA
(3)  No. of atoms in 2 g H2  = Moles *NA*atomicity
= 2/2 *NA * 2
= 2 NA
(4)  No. of atoms in 4 g He  = Moles *NA*atomicity
= 4/4 *NA *1
        = 1 NA
Thus Correct Ans is Option (c)
Anjali kumari
104 Points
5 years ago
The above answer (c) is correct. Siwa dosti atah dristi dristi they ye what I can fill to fulfill 100 characters.
Ahiri Basu
13 Points
5 years ago
Ammonia=4.25/17=1/4
=1/4×6.022×10^23×4
=6.022×10^23
Oxygen=8/32=1/4
=1/4×6.022×10^23×2
=3.011×10^23
Hydrogen=2/2=1
=6.022×10^23×2
=12.044×10^23
Helium=4/4=1
=6.022×10^23
Therefore,Hydrogen contains the highest no of atom.
 

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