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)

In in $1$ mole of $(N{H}_4{)}_{2}C{r}_2{O}_7$, the number of molecules $=6.023\times 10^{23}$
$\therefore$ In $1$ mole $(N{H}_4{)}_{2}C{r}_2{O}_7$, the total number of atoms of all elements present
$=6.023\times 10^{23}\times 19$
$=114.43\times 10^{23}$