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 $=\frac{the\,\,element\,\,or\,\,constituent}{mol.\,\,mass\,\,of\,\,the\,\,compound}\times 100$ 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 $viz\,\,NH_{4}^{+}$ $=2\times (14+4)=36$ $\therefore$ % of $NH_{4}^{+}=\frac{36}{252}\times 100$ $=14.29%$