Calculate the no. Of atoms 
A) 5.6 L of NH3 
B) 4.4g of Co2 
C) 52u of He atomic weight =4u)

Asked by rajdhan.yadav187350 | 12th Nov, 2018, 02:05: PM

Expert Answer:

A) 5.6 litre of NH3 
 
We space have comma

space 22.4 space litre space equals space 1 space mole space of space space NH 3 space space

therefore space 5.6 space litre space equals fraction numerator 5.6 cross times 1 over denominator 22.4 end fraction space mol

space space space space space space space space space space space space space space space space space equals space 0.25 space mol

1 space mol space of space gas space contains space 6.023 cross times 10 to the power of 23 space molecules

space 0.25 space mol space will space contain space equals space fraction numerator space 0.25 cross times 6.023 cross times 10 to the power of 23 over denominator 1 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.5 cross times 10 to the power of 23 space molecule

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4 cross times space 1.5 cross times 10 to the power of 23 space atoms

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 6.0 cross times 10 to the power of 23 space atoms space of space NH subscript 3
 
B) 4.4 g of CO2 
 
1 space mole space of space space CO subscript 2 space equals space 44 space straight g

4.4 space straight g space of space CO subscript 2 space equals fraction numerator 4.4 over denominator 44 end fraction

space space space space space space space space space space space space space space space space space space space space equals 0.1 space mole space of space CO subscript 2

1 space mole space of space CO subscript 2 space contains space 6.022 cross times 10 to the power of 23 space molecules

0.1 space mole space of space CO subscript 2 space will space contain space fraction numerator 0.1 cross times space 6.022 cross times 10 to the power of 23 over denominator 1 end fraction

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 6.022 cross times 10 to the power of 22 space molecules

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space 6.022 cross times 10 to the power of 22 space cross times space 3 space atoms

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1.80 space cross times 10 to the power of 23 space atoms space of space CO subscript 2
 
C)  52 u of He 

We space know comma

4 space space straight u space of space He space equals space 1 space atom space of space He

So space 52 space straight u space of space He space equals space 52 over 4 space

space space space space space space space space space space space space space space space space space space space space space space space equals space 13 space atoms space of space He
 
5.6 litre of NH3 contains 6×1023 atoms of NH3 

4.4 g of CO2 contains 1.8 ×1023 atoms of CO2
 
52 u of He contains 13 atoms of He.
 
 

Answered by Varsha | 13th Nov, 2018, 11:25: AM