Question

Arrange in increasing order of number of atoms in 1.6 g ${CH}_4$, 1.7 g ${NH}_3$ and 1.8 g $H_{2}O$.

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

Answer 1

Answer: (A)

$H_{2}O={NH}_3={CH}_4$

Solution:- (A)

As we know that,

$\text{No. of moles}\left(n\right)=\frac{\text{Given weight}}{\text{Mol. weight}}$

Therefore,

• Molecular weight of $C{H}_4=16g$
  

Given weight of $C{H}_4=1.6g$

$\therefore n_{C{H}_4}=\frac{1.6}{16}=0.1\text{mole}$

• Molecular weight of $N{H}_3=17g$
  

Given weight of $N{H}_3=1.7g$

$\therefore n_{N{H}_3}=\frac{1.7}{17}=0.1\text{mole}$

• Molecular weight of $H_{2}O=18g$
  

Given weight of $H_{2}O=1.8g$

$\therefore n_{H_{2}O}=\frac{1.8}{18}=0.1\text{mole}$

As we know that,

No. of atoms present in a molecule $=$ Avogadro's constant $\times$ No. of moles

$\because n_{C{H}_4}=n_{N{H}_3}=n_{H_{2}O}$

Therefore, No. of atoms are also equal.