$15 mL$ of a mixture of $N X_{2}$ and $NO$ was mixed with excess $O X_{2}$ and the $NO X_{2}$ formed was absorbed completely in alkali. The total volume decreased by $9 mL$ (all at the same $T$ and $P$ ). The volume of $N X_{2}$ in the mixture was

Answer: A

2 NO + O X_{2} 2 NO X_{2}

. For

v

mL of NO:

v

mL NO and

v /2

O X_{2}

disappear and

v

NO X_{2}

is formed and then absorbed. Net decrease

= v + v /2 = 1.5 v = 9

, so

v = 6 mL

NO.

N X_{2}

= 15 - 6 = 9 mL