The concentration of the alkane c20h42 (fm 282.55) in a particular sample of rainwater is 0.14 ppb. assume that the density of rainwater is close to 1.00 g/ml and find the molar concentration of c20h42.

The molar concentration is concentration in moles /Liter 1kg=1000g in a ltr. of water mass, So by assuming that the density of rainwater is close to 1.00 g/ml.the concentration of C20H42 (fm 282.55) in a particular sample of rainwater is 0.14 ppb is 0.32*10^-9*10^3= 0.32*10^-6g/L now, to have this in mole divide by molar weight 0.32*10^-6/282.55=1.133*10^-9mole/Liter. 
finally  
So answer 1.133*10^-9M

It changes food into energy that the organism can use.

2.00 is the !