Question

State true or false.

Number of atoms in $1.6g$ $C{H}_4$, $1.7g$ $N{H}_3$ and $1.8g$ $H_{2}O$ are equal.

• A. True
• B. False

Answer 1

The correct option is B False

The given statement is false.
True statement is, "Number of molecules in 1.6g $C{H}_4$, 1.7g $N{H}_3$ and 1.8 g $H_{2}O$ are equal".
The number of molecules $=\frac{\text{Weight}}{\text{Molecular weight}}\times \text{Avogadro's number}$
The number of molecules in 1.6 g of $C{H}_4=\frac{\text{1.6 g}}{\text{16 g/mol}}\times \text{N}=0.1N$
Note : Here, N is Avogadro's number.
The number of atoms in 1.6 g of $C{H}_4=5\times \text{the number of molecules}=5\times 0.1N=0.5N$
The number of molecules in 1.7 g of $N{H}_3=\frac{\text{1.7 g}}{\text{17 g/mol}}\times \text{N}=0.1N$

The number of atoms in 1.7 g of $N{H}_3=4\times \text{the number of molecules}=4\times 0.1N=0.4N$
The number of molecules in 1.8 g of $H_{2}O=\frac{\text{1.8 g}}{\text{18 g/mol}}\times \text{N}=0.1N$
The number of atoms in 1.8 g of $H_{2}O=3\times \text{the number of molecules}=3\times 0.1N=0.3N$