Answer :

(i) According to Avogadro’s Law,

1 mole of Argon 6.023×1023 atoms of Argon


So, 52 moles of Argon X atoms of Argon


X =52×6.023×1023


=3.13196×1025


Therefore in 52 moles of Argon 3.13196×1025 atoms are present.


(ii) Atomic Mass of Helium [He] = 4 u

This means that mass of one atom of Helium is = 4 u

So, Number of atom of Helium contained in 52 u = 52/4= 13 atoms

Therefore, in 52 u of Helium 13 atoms are present.


(iii) Atomic Mass of Helium [He] = 4 u

Number of Moles of Helium in 52 g = 52/4= 13 moles

So, Number of atoms in 52g of Helium [He] = 13×6.023×1023= 7.8299×1024


Therefore, in 52 g of Helium 7.83×1024 atoms are present.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

How much copper cNCERT - Chemistry Part-I

Calculate the numNCERT - Chemistry Part-I