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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`
2 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
```
2 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.
```
2 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 NAThus Correct Ans is Option (c)
```
2 years ago
```							The above answer (c) is correct. Siwa dosti atah dristi dristi they ye what I can fill to fulfill 100 characters.
```
2 years ago
```							Ammonia=4.25/17=1/4=1/4×6.022×10^23×4=6.022×10^23Oxygen=8/32=1/4=1/4×6.022×10^23×2=3.011×10^23Hydrogen=2/2=1=6.022×10^23×2=12.044×10^23Helium=4/4=1=6.022×10^23Therefore,Hydrogen contains the highest no of atom.
```
one year ago
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