# Question #b64d1

Jul 31, 2017

MW is 18 grams per mol (for water). In 27 grams of water there are $9.03 \times {10}^{23}$ molecules of water.

#### Explanation:

Water has molecular weight of 18 grams per mol (remember that hydrogen atomic weight is 1 and oxygen atomic weight is 16 grams and water formula is ${H}_{2} O$).

1 mole of water contains $6.02 \times {10}^{23}$ molecules.

If you have 27 grams of water, it means you have $\frac{27}{18} = 1.5$ moles of water.

If you have 1.5 moles of water, it means $1.5 \times 6.02 \times {10}^{23}$ water molecules are in it.

When you do the multiplication, you will get

1.5 moles of water contain $9.03 \times {10}^{23}$ molecules of water.

Jul 31, 2017

$9.03 \times {10}^{23} \text{Molecules}$

#### Explanation:

Recall $\to \text{no of moles" = "No of Entity"/"Avogadro Constant}$

Also $\to \text{no of moles" = "mass"/"molar mass}$

Water $\to {H}_{2} O$

$\text{Mass of} \textcolor{w h i t e}{x} {H}_{2} O \to 27 g$

$\text{Molar mass of} \textcolor{w h i t e}{x} {H}_{2} O \to \left(1 \times 2\right) + 16 = 2 + 16 = 18 g m o {l}^{-} 1$

$\therefore \text{no of moles of} \textcolor{w h i t e}{x} {H}_{2} O = \frac{27 \cancel{g}}{18 \cancel{g} m o {l}^{-} 1}$

$\text{no of moles of} \textcolor{w h i t e}{x} {H}_{2} O = 1.5 m o l s$

But $\text{No of Entity" = "no of moles" xx "Avogadro Constant}$

Hence $\text{No of Entity (Molecules)} = 1.5 \times 6.02 \times {10}^{23}$

$\text{No of Entity (Molecules)" = 9.03 xx 10^23 "Molecules}$