Question

A sample of ammonium phosphate $(N{H}_4{)}_{3}P{O}_4$ contains $3.18$ moles of $H-atom$. The number of moles of $O-atom$ in the sample is:

• A. $0.265$
• B. $0.795$
• C. $1.06$
• D. $3.18$

Answer 1

The correct option is C $1.06$

$149gm$ of $(N{H}_4{)}_{3}P{O}_4$ contains $12g$ of $H$.

For $3.18$ moles of $H$ i.e. $3.18g$ of $H$,

$3.18gm$ of $H$ present in $\frac{149\times 3.18}{12}=39.5gm(N{H}_4{)}_{3}P{O}_4$.

$149gm$ of $(N{H}_4{)}_{3}P{O}_4$ contains $64g$ of Oxygen.

$\therefore$ $39.5gm$ of $(N{H}_4{)}_{3}P{O}_4$ will contain $\frac{64\times 39.5}{149}g$ oxygen.

$=16.96gm$ oxygen

i.e. $16.96g$ oxygen present in the sample.

$\therefore$ No. of moles of oxygen=$\frac{16.96}{16}=1.06moles$

Hence, the correct option is $\text{C}$