Average diameters of colloids

[kasse]

A dispersion of spherical polymer particles contains Np = 4•1013 particles/dm3. The total surface of the particles in 1 dm3 is Asp = 820 m2.

a) What kind of average diameter can we calculate. Calculate this average diameter.

b) If we know that the dispersion contains 60 vol% water, we can calculate another type of average diameter. Which? Find it!

a) Surface average.

$$\bar{d_s} = \sqrt{\frac{\Sigma n_i d_i^2}{\Sigma n_i}}$$

b) Volume average.

$$\bar{d_s} = \sqrt{\frac{\Sigma n_i d_i^3}{\Sigma n_i}}$$

I don't know how to incorporate the data into the formulas.

 

[Simon_Moore]

Can't you just work it out by taking the surface area of a sphere and working back to get the radius?

You know the total surface area of the total particles in 1 litre, and you know how many particles there are in 1 litre, so from that you can work out the surface area of 1 particle.

 

[Blueshift5]

a) To calculate the surface average diameter, we can use the formula:

\bar{d_s} = \sqrt{\frac{6}{\pi N_p}\frac{Asp}{\Sigma n_i}}

Plugging in the values given, we get:

\bar{d_s} = \sqrt{\frac{6}{\pi (4 \cdot 10^{13})}\frac{820 \text{ m}^2}{1 \text{ dm}^3}} = 2.62 \times 10^{-7} \text{ m}

b) To calculate the volume average diameter, we can use the formula:

\bar{d_v} = \sqrt[3]{\frac{6}{\pi N_p}\frac{V}{\Sigma n_i}}

Since we know that the dispersion contains 60 vol% water, we can calculate the volume of water in 1 dm3 as:

V = 1 \text{ dm}^3 \cdot 0.60 = 0.6 \text{ dm}^3

Plugging in the values given, we get:

\bar{d_v} = \sqrt[3]{\frac{6}{\pi (4 \cdot 10^{13})}\frac{0.6 \text{ dm}^3}{1 \text{ dm}^3}} = 1.14 \times 10^{-7} \text{ m}

Therefore, the volume average diameter of the colloids in this dispersion is 1.14 nanometers.