Question

Percentage of cation in ammonium dichromate ${(N{H}_4)}_{2}C{r}_2{O}_7$ is

• A. $41.29$
• B. $14.29$
• C. $20.29$
• D. None of these

Answer 1

Answer: (B)

Percentage of the element or constituent no. of parts by mass $=\frac{theelementorconstituent}{mol.massofthecompound}\times 100$ Molecular formula of ammonium dichromate is ${(N{H}_4)}_{2}C{r}_2{O}_7$ mol. mass of ${(N{H}_4)}_{2}C{r}_2{O}_7$ $=2\times (14+4)+2\times 52\times 7\times 16=252$ No. of parts by mass of cation $vizN{H}_4^{+}$ $=2\times (14+4)=36$ $\therefore$ % of $N{H}_4^{+}=\frac{36}{252}\times 100$ $=14.29$