Question

How many grams of CaO2 are in 6.90x10^23 molecules of CaO2

Answer 1

Answer: 82.8 grams of $CaO_2$ are there in $6.90\times 10^{23}$ molecules of

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=\frac{6.90\times 10^{23}}{6.023\times 10^{23}}=1.15moles$

1 mole of $CaO_2$ weighs = 72 g

1.15 moles of $CaO_2$ weighs =$\frac{72}{1}\times 1.15=82.8g$

Thus 82.8 grams of $CaO_2$ are there in $6.90\times 10^{23}$ molecules of