Calculate normality of normal saline at 0.85%?
I take it that this is 0.85% w/v. If so, that means 0.85 g NaCl per 100 mL solution.
Convert 0.85 g NaCl to moles. moles = grams/molar mass = 0.85/58.5 (you may want to refine this figures) = 0.0145 moles/100 mL which is 0.145 moles/L soln which is 0.145 M or 0.145 N.
The normality is twice as the molarity because in this case normality is used in reference to tonicity (osmotic pressure) and NaCl splits in two ions doubling the number of osmotically active particles. It would be 2 x 0.145 x 0.93 = 0.270
0.93 is a correction factor for an imperfect solution of NaCl. If you obviate this factor the answer is 0.290 osmols which is normal serum osmolality. Called for this reason "Normal Saline" but it actually is 0.290Normal
0.145
To calculate the normality of a solution, we need to know the molarity and the number of equivalents of the solute.
In the case of normal saline, it is a 0.85% solution. This means that there are 0.85 grams of sodium chloride (NaCl) dissolved in 100 mL (or 100 grams) of water.
To find the molarity of the solution, we need to convert the percentage to grams per liter (g/L).
Step 1: Convert the percentage to grams
0.85% of 100 grams is (0.85/100) x 100 = 0.85 grams.
Step 2: Convert grams to moles
To convert grams of NaCl to moles, we use the molar mass of NaCl, which is approximately 58.5 g/mol.
Number of moles = Mass of NaCl / Molar mass of NaCl
Number of moles = 0.85 g / 58.5 g/mol ≈ 0.0145 mol
Step 3: Convert mL to L
Since we want to find the normality in terms of liters, we need to convert the volume from milliliters (mL) to liters (L).
Volume in L = Volume in mL / 1000
Volume in L = 100 mL / 1000 = 0.1 L
Step 4: Calculate molarity
Molarity (M) = Number of moles / Volume in L
Molarity (M) = 0.0145 mol / 0.1 L = 0.145 M
Now, we have the molarity of the solution, which is 0.145 M.
Next, we need to determine the number of equivalents of the solute. In the case of NaCl, it is a 1:1 electrolyte, meaning that one mole of NaCl dissociates into one mole of Na+ ions and one mole of Cl- ions. Therefore, the number of equivalents is equal to the number of moles.
Number of equivalents = Number of moles = 0.0145 mol
Finally, the normality is equal to the number of equivalents divided by the volume in liters.
Normality = Number of equivalents / Volume in L
Normality = 0.0145 mol / 0.1 L ≈ 0.145 N
Hence, the normality of normal saline at 0.85% is approximately 0.145 N.