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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
one year 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)
one year 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.
one year 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)
one year ago

The above answer (c) is correct. Siwa dosti atah dristi dristi they ye what I can fill to fulfill 100 characters.
one year 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.

6 months ago
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