Weight specific surface area?

Question

Calculate the weight specific surface area of 10 g lactose powder (powder density = 0.75g/cm^3). Assume that the lactose particles are non-porous spheres with a mean diameter of 100 um.

I don't understand exactly what I am meant to calculate, the surface area of each particle? How does the weight of the powder as a whole come into the picture in that case?

Many thanks

Answer 1

1] Calculate the volume and area of a 100 micron sphere

2] Calculate the mass of one sphere with the density given

3] Calculate how many spheres are required to give 1 g

4] Calculate the area of that many spheres

by "Weight specific surface area" I think they want something like cm^2/g

Answer 2

Specific Surface Area Of Sphere

Answer 3

The 0.75 g/cm^3 is the volume-specific density of the powder. So the volume of any 100 um

sphere divided by its mass (m) would be this value.

The surface-area-specific value (call it Sa) will be the surface area of the 100 um sphere with the same mass (m) for the sphere as used in the given volume-specific density.

Volume of the sphere/w = 0.75 g/cm^3 , so that m = (0.75 g/cm^3)/volume

Now

Sa = [4*pi*D^2/4] since Sa = 4*pi*radius squared = 4*pi*(D/2)^2, with D = diameter

But Sa/m value is what we want, and we don't know what value m has.

So use the 0.75 g/cm^3 multiplied by the volume of the sphere to get the mass of each

molecule (on average)

V = (4/3)*pi*(D^3/8) in cm^3 where D = diameter, must be entered in cm units (not just 100 um).

because (radius^3) = (D^3)/2^3 =(D^3)/8 , and now

m = [0.75 g/cm^3]*V in grams for m

And thus

Sa/w = [4*pi*D^2/4]/m = [pi*D^2]/[(4/3)*pi*(D^3/8)*(0.75 g/cm^30] {put in 3/4 for the0.75 to

cancel the 4/3 factor}

Now just substitute the cm value (of 100 um) for D in this last expression to calculate the answer.

Answer 4

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