Question

Water has a mass per mole of 18.0 g/mol, and each water molecule (h2o) has 10 electrons. (a) how many electrons are there in 7.74 liters of water? (1 liter is equivalent to 1.00 × 10-3 m3) (b) what is the net charge of all these electrons?

Answer 1

Answer: a) $2.58\times 10^{27}$ electrons are present is 7.74 L of water.

b)$2.7\times 10^{8} Coulombs$ is the net charge of all these electrons

Explanation:

a) Molar mass of the water = 18 g/mol

Volume of water = 7.74 L

Density of water = 1000 g/L

Mass of water in 7.74 L of water :

$density=1000 g/L=\frac{mass}{volume}=\frac{mass}{7.74 L}$

Mass of the water = 7740 g

Moles of water =$\frac{\text{mass of water}}{\text{molar mass of water}}=\frac{7740 g}{18 g/mol}=430 moles$

$1 mol=6.022\times 10^{23}$molecules

Number of water molecules in 430 moles:

$430\times 6.022\times 10^{23} =2.58\times 10^{26}$molecules

1 molecule water contains = 10 electrons

Then$2.58\times 10^{26} molecules=2.58\times 10^{26}\times 10=2.58\times 10^{27}$ electrons

$2.58\times 10^{27}$ electrons are present is 7.74 L of water.

b) The net charge of all these $2.58\times 10^{27}$ electrons:

Charge on 1 electron =$-1.602\times 10^{-19} Coulombs$

The net charge of all these $2.58\times 10^{27}$ electrons:

$2.58\times 10^{27}\times 1.602\times 10^{-19} C=2.7\times 10^{8} Coulombs$

$2.7\times 10^{8} Coulombs$ is the net charge of all these electrons