# Question #e845d

Apr 2, 2017

$27.1 g$ of $C a O$ contains $2.92$ x ${10}^{23}$molecules.

#### Explanation:

This is a problem of stoichiometry.

The molar mass of $C a O = 40 + 16$grams, i.e. $56$ grams.

Now, the Avogadro number $= 6.023 x {10}^{23}$.

So, $56 g$ of $C a O$ contains $6.023$ x ${10}^{23}$ molecules.

$\therefore 27.1 g$ of $C a O$ contains $2.92$ x ${10}^{23}$molecules. (Answer).

Hope it Helps :)