Question

What is number of moles present in $126$ amu of ${HNO}_3$?

• A. $2$
• B. $\frac{2}{N_A}$
• C. $6$
• D. $\frac{6}{N_A}$

Answer 1

Answer: (A)

$2$

$1$ mole of $HN{O}_3=1+14+(3\times 16)$

$\Rightarrow 1+14+48=63g=63amu$

$\therefore 126amu$ of $HN{O}_3=\frac{126}{63}=2$ moles