Find the Molality, Millimolality, and Milliosmolality of both Solutions.

I found this on another website:

Molality = moles of solute/kilogram of solvent.

Osmolality = osmoles of solute/kilogram of solvent. (An osmole is the number of moles that contribute an osmotic pressure.)

Also, milli is 10^-3

For the first solution, start by calculating the number of moles of NaCl in 1.95g.  n=m/M= 1.95g / 58.44g/mol = 0.033 mol.

0.033 mol per 250mL of water, is the same as 0.033 x 4 = 0.133 mol per litre of water.

So the molality is 0.133 M.

'milli' is 10^-3.  For example, a millimeter is 1/1000 (which is 10^-3) of a meter.  So the millimolality of the solution will be 133 mM.

Osmolality is slightly more complex.  It takes into account that a molecule may dissociate (split apart) in solution. 

For the first solution, the milliosmolality is simply the millimolality (133mM) multiplied by 2 (since NaCl dissociates into 2 ions in solution: Na+ and Cl-).

Glucose, however, does not dissociate, so in the case of solution 2 the millimolality and milliosmolality will be the same.

Hey Kelly,

I found an excellent resource for you that explains this topic to the bone. I attached it as a PDF file below. Take a read at it and follow it. You'll find your question (with some variation) in there.

Reply here if you need further assistance; it's been a long time since I've looked at this so I may need some review as well. I am a biology teacher so it somewhat relates to what I'm teaching.