[shaina2006]

I need help wit dis question! Please help me!!!

The NaCl solution has a concentration of 1.95 g per 250 ml (molecular weight= 58.5). The glucose solution has a concentration of 9.0 g per 250 ml (molecular weight= 180). Calculate the molality, millimolality, and milliosmolality of both solutions.

[Sev]

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.)
http://en.wikipedia.org/wiki/Osmolality

Also, milli is 10^-3

[shaina2006]

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.)
http://en.wikipedia.org/wiki/Osmolality

Also, milli is 10^-3

could u please show me how 2 do it? I have no clue wat 2 do!!

[Sev]

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

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.