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% $