Question

The total number of protons present in $124g$ of $N{O}_3^{-}$ are:

• A. $124$
• B. $62N_A$
• C. $31N_A$
• D. $62$

Answer 1

Answer: (B)

Protons in $N{O}_3^{-}=31$
$Moles$ of $N{O}_3^{-}=\frac{124}{62}=2mole$
$1moleN{O}_3^{-}=31N_A$ protons
$2moleN{O}_3^{-}=62N_A$