Explanation:
The given data is as follows.
Concentration = 0.1
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
=
T = = (30 + 273) K = 303 K
Formula for electric double layer thickness () is as follows.
=
where, = concentration =
Hence, putting the given values into the above equation as follows.
=
=
= m
or, =
= 1 nm (approx)
Also, it is known that =
Hence, we can conclude that addition of 0.1 of KCl in 0.1
of NaBr "
" will decrease but not significantly.
