You are.
mols NaCl = 0.9g/58.44 = ?
?/0.1 = M
I'm doing something wrong with osmolarity. I have 100 ml of .9% NaCl.
So what i did was take the MW (58.44) and divided it by the .9 because it should be grams, then divided that number by .1 (for the liters) but its coming out with a huge number and its not supposed to. I feel like im doing this completely backwards, help??
3 answers
what if you were mixing two different solutions? such as .9% of the NaCl solution and a 5% dextrose solution? would you just find the moles of both like the above and add them together? then divide all of it by the L?
Along those lines, yes. The calculation is complicated by the fact that NaCl ionizes completely into two ions while the dextrose isn't ionized at all.
With NaCl you must use
pi = i*nRT where i is 2 for NaCl.
For dextrose i = 1.
So I would calculate M NaCl and M dextrose, find %NaCl and % dextrose, then
pi = i*nRT*(%NaCl) + i*nRT(%dextrose). Check my thinking.
With NaCl you must use
pi = i*nRT where i is 2 for NaCl.
For dextrose i = 1.
So I would calculate M NaCl and M dextrose, find %NaCl and % dextrose, then
pi = i*nRT*(%NaCl) + i*nRT(%dextrose). Check my thinking.
2 answers