Question

Density of $H_{2}O$ is $1gm{L}^{-1}$. If we have $1mL{H}_{2}O$, then:

• A. it has $55.55mol{L}^{-1}$
• B. it has $3.33\times {10}^{22}{H}_{2}Omolecules$
• C. it has $3.33\times {10}^{22}Hatoms$
• D. it has $6.66\times {10}^{22}Oatoms$

Answer 1

Answer: (A), (B)

The correct options are
A it has $55.55mol{L}^{-1}$
B it has $3.33\times {10}^{22}{H}_{2}Omolecules$

Density of $H_{2}O=1gm{L}^{-1}=1000g{L}^{-1}$

$=\frac{1000}{18}=55.55mol{L}^{-1}$

$1g=\frac{1}{18}mol=\frac{6.02\times {10}^{23}}{18}{H}_{2}Omolecules$

It has $6.06\times {10}^{22}Hatoms$.

It has $3.33\times {10}^{22}Oatoms$.

Hence, options $A$ and $B$ are correct.