Question

The total number of atoms of all elements present in $1$ mole of ammonium dichromate is

• A. $19$
• B. $6.023 \times 10^{23}$
• C. $114.437 \times 10^{23}$
• D. $84.322 \times 10^{23}$

Answer 1

Answer: (C)

$114.437 \times 10^{23}$

Answer 2

In in $1$ mole of $(NH_{4})_{2} Cr_{2}O_{7}$, the number of molecules $=6.023\times 10^{23}$ $\therefore$ In $1$ mole $(NH_{4})_{2} Cr_{2}O_{7}$, the total number of atoms of all elements present $=6.023\times 10^{23}\times19$ $=114.43\times 10^{23}$