Calculate the total number of moles of particles (Na+ and Cl-) in the evaporating dish?

Question

Info:

Molar Mass of NaCl = 58.44g

Mass of NaCl solution = 1.478g

Solution = 0.9% NaCl per 100 mL

Molarity of Saturated NaCl Solution = 2.526 Mol/L

Questions:

1) Calculate the total number of moles of particles (Na+ and Cl-) in the evaporating dish?

2) Calculate the osmolarity of the saturated NaCl solution.

3) What is the osmolarity of the 0.9% Sodium Chloride IV solution bag?

Answer 1

Well, this is a fun problem! I think what is tripping you up is the vocabulary, so let's try an analogy:

There are 12 doughnuts in a dozen and they weigh 100 grams all together

There are 6.022 x 10²³ atoms in a mole and it weighs (molar mass) x grams

Molar Mass of NaCl = 58.44g This how much it weighs if you had 6.022 x 10²³ molecules of NaCl

Mass of NaCl solution = 1.478g The is how much the solution you have weighs

Solution = 0.9% NaCl per 100 mL This is the ratio (percentage) of actual NaCl molecules per 100 ml of solution.

Molarity of Saturated NaCl Solution = 2.526 Mol/L How many mols are in a liter

Question: Calculate the total number of moles of particles (Na+ and Cl-) in the evaporating dish?

1) Figure out the ratio of Na and Cl. 22.9898/35.453 g

2) Figure out how much NaCl is in the actual solution: 0.9% of 100ml for 1.478g (1 gram = 1 ml)

0.9% of 1.478g = 0.000133

3) This is equal to 0.000133mol of NaCl; so how many total molecules is: 0.000133(6.022 x 10²³ )= 8.010 X 10¹⁹

4) Now we know there is a ratio of Na to Cl of 22.9898/35.453 g or Na is about 39% and Cl is 61%

Na= 39% of 8.010 X 10¹⁹= 3.124 X 10¹⁹ atoms

Cl = 61% of 8.010 X 10¹⁹= 4.890 X 10¹⁹ atoms