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Question #e845d

1 Answer

Aditya Banerjee.

Apr 2, 2017

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

Explanation:

This is a problem of stoichiometry.

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

Now, the Avogadro number $= 6.023 x 10^{23}$ $=6.023x10^23$ .

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

$:.27.1g$ of $CaO$ contains $2.92$ x $10^23$ molecules. (Answer).

**Hope it Helps :) **

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