Question

The density of water at $4^\circ C$ is $1.0 \times {10^3}kg{m^{ - 3}}$. The volume occupied by one molecule of water is approximately:
A. $3.0 \times {10^{ - 23}}mL$
B. $6.0 \times {10^{ - 22}}mL$
C. $3.0 \times {10^{ - 21}}mL$
D. $9.0 \times {10^{ - 23}}mL$

Answer 1

Hint: We can use the relationship between the density, mass and volume to determine one of these when other two are given.

Complete step by step answer:
We can convert the volume in ${m^{ - 3}}$ to $c{m^{ - 3}}$by using the following conversion factor:
$\left( {\dfrac{{1\;{m^3}}}{{{{10}^6}\;c{m^3}}}} \right)$
The volume in $c{m^{ - 3}}$ can be further converted to mL by the following conversion factor:
$\left( {\dfrac{{1\;c{m^3}}}{{1\;mL}}} \right)$
We can convert the units of given density by using the above conversion factors for volume as follows:

Mass in $kg$ can be converted to grams by the following conversion factor:
$\left( {\dfrac{{1\;kg}}{{1000\;g}}} \right)$
Now, we can convert the units of density by using the above conversion factors for mass as follows:

Now, we have to calculate the mass of one molecule of water. In order to do that, we will use the molar mass of water and the Avogadro constant whose values are $M\left( {{\rm{water}}} \right) = 18.0\;gmo{l^{ - 1}}$ and ${N_A} = 6.0 \times {10^{23}}\;mo{l^{ - 1}}$ respectively.
We know that molar mass of water is the mass of one mole of water and the number of particles in one mole of water is given by the Avogadro constant. From these two, we can infer that one mole of water weighing $18.0\;g$ contains $6.0 \times {10^{23}}$ water molecules. It means that mass of $6.0 \times {10^{23}}$molecules is $18.0\;g$. Let’s calculate the mass of one molecule of water as follows:
$\dfrac{{18.0\;g}}{{6.0 \times {{10}^{23}}{\rm{ molecules}}}} = 3.0 \times {10^{ - 23}}{\rm{ }}g{\rm{molecul}}{{\rm{e}}^{ - 1}}$
Finally, we can calculate the volume of one molecule of water by using its mass and density as follows:
$\begin{array}{c}
density = \dfrac{{mass}}{{Volume}}\\
Volume = \dfrac{{mass}}{{density}}\\
 = \dfrac{{3.0 \times {{10}^{ - 23}}{\rm{ }}g}}{{1.0\;gm{L^{ - 1}}}}\\
 = 3.0 \times {10^{ - 23}}{\rm{ }}mL
\end{array}$
Hence, the volume of one molecule of water at $4^\circ C$ is $3.0 \times {10^{ - 23}}{\rm{ }}mL$

Hence, the correct option is (A).

Note:
We have to be very careful with the units for each conversion and calculation.