Question

Calculate the number of atoms of each element in 31.50 g of $HN{O}_3$.
($H=1;N=14;O=16$)

• A. $3.011\times {10}^{23}$ of $H$, $3.011\times {10}^{23}$ of $N$, $9.033\times {10}^{23}$ of $O$
• B. $6.022\times {10}^{23}$ of $H$, $3.011\times {10}^{23}$ of $N$, $9.033\times {10}^{23}$ of $O$
• C. $3.011\times {10}^{23}$ of $H$, $6.022\times {10}^{23}$ of $N$, $9.033\times {10}^{23}$ of $O$
• D. $3.011\times {10}^{23}$ of $H$, $3.011\times {10}^{23}$ of $N$, $6.022\times {10}^{23}$ of $O$

Answer 1

Answer: (A)

$3.011\times {10}^{23}$ of $H$, $3.011\times {10}^{23}$ of $N$, $9.033\times {10}^{23}$ of $O$

given in this question 31.5gm of $HN{O}_3$

so the mol of $HN{O}_3$ is $\frac{31.5}{63}$ = 0.5 mol

then mol of each atom is

H is 0.5 mol mean $0.5\times 6.022\times {10}^{23}$ no of atom

N is 0.5 mol mean $0.5\times 6.022\times {10}^{23}$ no of atom

and O is 1.5 mol $1.5\times 6.022\times {10}^{23}$ no of atom

so the final answer is A