Question

What is the average volume occupied by $1$ molecule of water $\left(H_{2}O\right)$?

• A. $3.99\times {10}^{-23}c{m}^3$
• B. $2.99\times {10}^{23}c{m}^3$
• C. $2.99\times {10}^{-23}c{m}^3$
• D. $4.99\times {10}^{-23}c{m}^3$

Answer 1

Answer: (B)

$2.99\times {10}^{23}c{m}^3$

To find the volume occupied by one molecule of $H_{2}O$,

we'll have to first convert the moles to grams.

Note that 1 mole is different from 1 1molecule.

1 mole of the compound contains Avogadro's number of molecules, atoms, small entities.

We know that,

1 mol of $H_{2}O$ = $6.0223\times {10}^{23}molecules$

1 molecule of $H_{2}O$ = $\frac{1mol{H}_{2}O}{6.0223\times {10}^{23}}$ $=1.66\times {10}^{-24}mol{H}_{2}O$

Number of moles = $\frac{\text{Weight taken}}{\text{Molecular Weight}}$

$\Rightarrow$ $1.66\times {10}^{-24}mol=\frac{\text{Weight taken}}{18}$

$\Rightarrow$ $Weight=18\times 1.66\times {10}^{-24}$

$=2.99\times {10}^{-23}g{H}_{2}O$

As we know that 1g of water $\left(H_{2}O\right)$ occupies $1cm^{3}

$2.99\times {10}^{-23}g$ of $H_{2}O$ occupies $2.99\times {10}^{-23}c{m}^3$