chloroalkanes , fluoroalkanes, etc do they all form hydrogen bonds with one another and only with water? what about the dispersion forces since they're hydrocarbons?
I didn't think chloroalkanes and fluoroalkanes can form hydrogen bonds with each other as there isn't any H bonded to F,O,N for them to form hydrogen bonds with. I believe that since they are polar, they would form dipole-dipole bonds if anything? If a chloroalkane or fluoroalkane has a long non-polar end, I suppose the net dispersion forces will be the strongest force of attraction between molecules.
But yes they do form hydrogen bonds with water as the H's are bonded to an O, allowing for hydrogen bonding to occur.
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Another question I have is in relation to calibration curves;
1 method of doing these questions is to use "dilation factors" to calculate the concentration in the sample. But I find its more logically to calculate the mass in the diluted sample and work backwards from here. Is this an acceptable method of working for these type of questions? 
I think you mean dilution factor? hahah
Well yeah i just ratios so much easier - im assuming this is what you do? so you have to fractions and you gotta solve for x?..
Yeah, I do ratios for some of those questions because it makes more sense to me.

What I meant was for instance, you determine c(Pb) in 10ml is = 20.0mg/L
So m(Pb) in 10ml is = 0.2mg
It's a 10ml aliquot from the 100ml flask it originally come from;
m(Pb) in 100ml = 2.0mg
And you're asked to find the concentration of Pb in a 2.0g sample.
% Pb = 2.0 * 10^3 g / 2.0g * 100 = 0.1%
Rather than doing that messy dilution factor some of these practice exams go on about. :S