Question

Number of atoms present in $1.8g$ $H_{2}O$, $1.7g$ $N{H}_3$ and $1.6g$ $C{H}_4$ in the increasing order is :

• A. $H_{2}O<N{H}_3<C{H}_4$
• B. $C{H}_4<N{H}_3<H_{2}O$
• C. $C{H}_4=N{H}_3=H_{2}O$
• D. $N{H}_3<C{H}_3<H_{2}O$

Answer 1

The correct option is A $H_{2}O<N{H}_3<C{H}_4$

Number of atoms present in,
$1.8g$ $H_{2}O$ is $0.1\times 3$ mol.
$1.7g$ $N{H}_3$ is $0.1\times 4$ mol.
and $1.6g$ $C{H}_4$ is $0.1\times 5$ mol.