Question

Number of atoms present in $52\text{g}$ of helium is:
$(N_A$ is Avogadro’s constant, Gram atomic mass of helium is $4\text{g})$​

• A. $13N_A$
  
• B. $N_A$
• C. $26N_A$
• D. $52N_A$

Answer 1

The correct option is A $13N_A$


Gram atomic mass of helium is $4\text{g}$ so number of atoms present in $52\text{g}$ of helium is
$\frac{52}{4}\times N_A=13N_A$