# Unit problem with calculating the mass concentration of a particle

1. Nov 11, 2011

### guitarstorm

1. The problem statement, all variables and given/known data

We have to calculate the mass concentration for a particle size distribution. I'm just doing the first size range here, since once I get that the rest is just repetition.

FYI, the particles are spherical.

Size range (diameter) = 0.005-0.5 μm
Average particle size (diameter) = 0.25 μm
Number concentration = 1 cm$^{-3}$
Particle density = 1800 kg m$^{-3}$

The answer is supposed to be in units μg m$^{-3}$ μm$^{-1}$

2. Relevant equations

$m=\rho _{p}\frac{\pi }{6}D_{p}^{3}n_{n}(D_{p})$

3. The attempt at a solution

First thing I did was convert the density from kg m$^{-3}$ to μg m$^{-3}$:

$\frac{1800 \: kg}{m^{3}} * \frac{1\: \mu g}{1*10^{9\: }kg}=\frac{1.8*10^{12}\: \mu g}{m^{3}}$

Then I plugged into the mass concentration formula:

$(1.8*10^{12}\mu g\: m^{-3})(\frac{\pi }{6})(0.25\: \mu m)^{3}(1\: cm^{-3})=1.473*10^{10}\mu g\: m^{-3\: }\mu m^{3}\: cm^{-3}$

So now, since there's no cm in the units we're supposed to have, I figured I'd get rid of that:

$(1.473*10^{10}\: \frac{\mu g\: \mu m^{3}}{m^{3}\: cm^{3}})*(\frac{100\: cm}{1\: m})^{3}=1.473*10^{16}\: \frac{\mu g\: \mu m^{3}}{m^{6}}$

So now we have this mess, and at this point I give up, because idk how the heck to get this in μg m$^{-3}$ μm$^{-1}$... Not to mention the number I'm getting seems way too large...