Respuesta :
Answer:
[tex]m=5.78\times 10^{-3}\ g[/tex]
[tex]n_e=1.935\times 10^{20}[/tex] is the no. of electrons
Explanation:
Given:
- quantity of charge transferred, [tex]Q=31\ C[/tex]
No. of electrons in the given amount of charge:
As we have charge on one electron [tex]1.602\times 10^{-19}\ C[/tex]
so,
[tex]n_e=\frac{Q}{e}[/tex]
[tex]n_e=\frac{31}{1.602\times 10^{-19}}[/tex]
[tex]n_e=1.935\times 10^{20}[/tex] is the no. of electrons
- Now if each water molecules donates one electron:
Then we require [tex]n=1.935\times 10^{20}[/tex] molecules.
Now the no. of moles in this many molecules:
[tex]n_m=\frac{n}{N_A}[/tex]
where
[tex]N_A=6.022\times 10^{23}[/tex] Avogadro No.
[tex]n_m=\frac{1.935\times 10^{20}}{6.022\times 10^{23}}[/tex]
[tex]n_m=3.213\times 10^{-4}\ moles[/tex]
- We have molecular mass of water as M=18 g/mol.
So, the mass of water in the obtained moles:
[tex]n_m=\frac{m}{M}[/tex]
where:
m = mass in gram
[tex]3.213\times 10^{-4}=\frac{m}{18}[/tex]
[tex]m=5.78\times 10^{-3}\ g[/tex]
