Question

How many molecules of oxygen are produced when a sample of 38.9 g of water is decomposed by electricity?

Answer 1

Answer:

A) $6.5\times 10^{23}\ \text{molecules}$

Explanation:

m = Mass of water = 38.9

M = Molar mass of water = 18 g/mol

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

The reaction of electrolysis would be

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$

Number of moles of $H_2O$

$n=\dfrac{m}{M}\\\Rightarrow n=\dfrac{38.9}{18}\ \text{mol}$

From the reaction it can be seen that 2 moles of $H_2O$ gives 1 mole of $O_2$

So, number of moles of $O_2$ produced is

$\dfrac{38.9}{18}\times \dfrac{1}{2}=1.081\ \text{mol}$

Number of molecules

$1.081N_A=1.081\times 6.022\times 10^{23}\\ =6.5\times 10^{23}$

So, $6.5\times 10^{23}\ \text{molecules}$ of oxygen is produced.