# How many water molecules in a 0.90*g mass of water?

Oct 25, 2017

$3.01 \times {10}^{22}$ molecules.

#### Explanation:

The mass of one mole of water is $\text{18 g}$. Therefore, $\text{0.9 g}$ of water is

$\text{0.9 g"/"18 g/mol" = "0.05 moles}$

of water. One mole of water contains Avagadro's constant, ie, $6.022 \times {10}^{23}$ molecules of water, therefore, $0.05$ moles of water contain

"0.05 mol" xx (6.022^23"molec.")/("1 mol") = 3.01xx10^22 $\text{molecules}$

Oct 25, 2017

We find the molar quantity.....and get $\text{number of water molecules}$ $= 3.01 \times {10}^{22}$...

#### Explanation:

$\text{Moles of water} = \frac{0.9 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 0.050 \cdot m o l$

And we know that there are $6.022 \times {10}^{23}$ molecules in one mole of any substance.....

And so we take the product......

$0.050 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 = 3.01 \times {10}^{22}$ $\text{water molecules}$. (And thus the product gives a dimensionless number as required....)

(i) how many oxygen atoms in this molar quantity; and (ii) how many hydrogen atoms......?

Oct 25, 2017

$30.1$ x ${10}^{21} m l c s$

#### Explanation:

$M a s s o f w a t e r \left({H}_{2} O\right)$ $= 0.9 g r a m s$

$M o l a r m a s s o f \left({H}_{2} O\right)$ $= 1.008$ x $2$ + 1$6$
$= 18.016 g m o {l}^{-}$

Number of molecules = Mass in grams$/$Molar mass x ${N}_{A}$

$= \frac{0.9}{18.016}$ x $6.02$ x ${10}^{23}$

$= 30.1$ x ${10}^{21}$ $m l c s$