# How many water molecules are present in 54 gram of water?

Jun 25, 2018

We get $1.81 \times {10}^{24}$ water molecules...

#### Explanation:

We assess the molar quantity of water in the usual way...

$\text{Number of moles"="mass"/"molar mass}$

$\equiv \frac{54 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 3.00 \cdot m o l$

But the $m o l$ is like the $\text{dozen}$, or the $\text{gross}$, it specifies a NUMBER, here ${N}_{A} \equiv 6.022 \times {10}^{23} \cdot m o l$. Why do we use such an absurdly large number? Because ${N}_{A} \cdot \text{^1H*"atoms}$ have a MASS of $1 \cdot g$ precisely...and thus the mole is the link between the micro world of atoms, and molecules, to the macro world of grams and litres....

And so (finally!) we take the product...

$\text{Number of water molecules"="molar quantity"xx"Avogadro's number.}$

$= {\underbrace{3.00 \cdot \cancel{m o l} \times 6.022 \times {10}^{23} \cdot \cancel{m o {l}^{-} 1}}}_{\text{a number as required.."=??*"water molecules}}$

Jun 25, 2018

$1.8 \times {10}^{24}$ water molecules.

#### Explanation:

There are $6.022 \times {10}^{23}$ molecules in a mole.

There are $18.01528$ grams of water per mole of water.

These can be figured out using the concept of moles and molecular weight. I have attached a video that provides a good explanation.

Explanation of moles and molecular weight

Now you have all the information needed to solve the problem.

You start out with 54 grams of water. You want to convert that into molecules, but how can you do that? You can't go directly from grams to molecules because there is no easy unit conversion. However, you can go from grams to moles using

$18.01528$ grams of water per mole of water.

From there, you can go from moles to molecules using $6.022 \times {10}^{23}$ molecules in a mole.

A T chart is useful for this kind of problem.

"54 g H"_2"O" xx ("1 mol H"_2"O")/("18.0152 g H"_2"O") xx(6.022 xx 10^23 \ "molecules")/("1 mol H"_2"O")

"54" cancel("g H"_2"O") xx (cancel("1 mol H"_2"O"))/(18.0152cancel ("g H"_2"O")) xx (6.022 xx 10^23 \ "molecules")/(cancel("1 mol H"_2"O"))

Notice when you apply this T chart, the units cancel out, leaving molecules in the final answer, which is the correct units.

Now actually multiply everything out

$\frac{54 \times 6.022 \times {10}^{23}}{18.0152} = 1.805 \times {10}^{24}$

You are left with $1.8 \times {10}^{24}$ water molecules (rounded to two sig figs).