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kristadwsn
23.08.2019 •
Chemistry
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) csci electrolyte solution at 30 °c. how are these numbers affected by the addition of 0.1 mol/dm3 of kcl? at what distance from the particle surface (r) has the potential decayed to 1% of its initial value?
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Ответ:
Explanation:
The given data is as follows.
Concentration = 0.1![mol/dm^{3}](/tpl/images/0191/8730/0b26c.png)
= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions
=![6.022 \times 10^{25} ions/m^{3}](/tpl/images/0191/8730/88437.png)
T =
= (30 + 273) K = 303 K
Formula for electric double layer thickness (
) is as follows.
where,
= concentration = ![6.022 \times 10^{25} ions/m^{3}](/tpl/images/0191/8730/88437.png)
Hence, putting the given values into the above equation as follows.
=
=
m
or, =![9.7 A^{o}](/tpl/images/0191/8730/424fa.png)
= 1 nm (approx)
Also, it is known that
= ![\sqrt \frac{1}{n^{o}}](/tpl/images/0191/8730/3e6da.png)
Hence, we can conclude that addition of 0.1
of KCl in 0.1
of NaBr "
" will decrease but not significantly.
Ответ:
Answer :
In my opinion , answer is A