# Question b7de9

Jul 16, 2017

There are approximately $6.0 \cdot {10}^{21}$ water molecules in $0.18 \text{ mL}$ of water.

#### Explanation:

This depends on the density of water, which depends on the temperature and pressure. Let's assume that the density is

$1.0 {\text{ ""g"*"mL}}^{-} 1$

This is technically only true at 4 ˚C# but is ok for a simplification.

The formula for density of any substance is:

$\text{density"="mass"/"volume}$

Rearrange this to solve for mass:

$\text{mass of water"="density"*"volume"=1.0*0.18=0.18" } g$

Now we need the number of particles, which chemists measure in moles. To find the number of moles, we need the mass and the molar mass. The molar mass is given by the sum of the individual molar masses of the constituent atoms in the molecule. In this case we have two hydrogens and one oxygen:

${\text{molar mass"=O+H+H=16+1.0+1.0=18" "g*"mol}}^{-} 1$

$\text{number of moles"="mass"/"molar mass"=0.18/18=0.01=10^-2" } m o l$

The Avogadro constant is the number of particles in one mole:

$\text{Avogadro constant} = 6.0 \cdot {10}^{23}$

The number of moles multiplied by the Avogadro constant will tell you how many molecules there are:

$\text{number of molecules"="number of moles"*"Avogadro constant} = {10}^{-} 2 \cdot 6.0 \cdot {10}^{23} = 6.0 \cdot {10}^{21}$

Another way of doing this would be to use the molar volume of water.