Question

The percent of $N$ in $66\%$ pure $(N{H}_4{)}_{2}S{O}_4$ sample is:

• A. $32$
• B. $28$
• C. $14$
• D. none of these

Answer 1

The correct option is C $14$

Molecular mass of $(N{H}_4{)}_{2}S{O}_4=132g$

$\%$ of Nitrogen $=\frac{2\times 14}{132}=21.21\%$

$100\%$ pure $(N{H}_4{)}_{2}S{O}_4$ sample contains $21.21\%$ of $N$

$\therefore 66\%$ pure $(N{H}_4{)}_{2}S{O}_4$ sample contains $=\frac{28}{132}\times 66$

$=14\%$ of $N$