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??
You are.
mols NaCl = 0.9g/58.44 = ?
?/0.1 = M
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.
To calculate osmolarity, you need to convert the concentration of solution from percentage to molarity. Let's break down the steps to correctly calculate the osmolarity of your 100 ml solution of 0.9% NaCl.
Step 1: Convert the given percentage to a fraction:
0.9% can be written as 0.009 (0.9 ÷ 100).
Step 2: Convert the fraction to molarity (moles per liter):
To convert from grams to moles, you need to know the molecular weight (MW) of NaCl, which is 58.44 g/mol.
Moles of NaCl = (0.009 g NaCl) / (58.44 g/mol) = 0.0001542 mol NaCl
Step 3: Calculate the osmolarity:
Now, you need to calculate the osmolarity using the number of moles of solute (NaCl) and the volume of the solution.
Osmolarity = (moles of solute) / (volume of solution in liters)
Volume of solution = 100 ml = 0.1 L (since 1 L = 1000 ml)
Osmolarity = (0.0001542 mol) / (0.1 L) = 0.001542 M
So, the osmolarity of your 100 ml solution of 0.9% NaCl is 0.001542 M.