Question

If a solution of 50% (w/v) NaCl was to be diluted to 100 mM, using water, you must use 1 part of 50% NaCl-solution and....... parts of water

The correct answer should be around 84,6 parts of water. Please explain how I can calculate this

Answer 1

A 50% (w/v) solution means if we take the volume of a solution to be 100 mL, then mass of solute in it = 50 g

Now lets say we have 100 mL of water

mass of solute (NaCl) in it = 50 g

Then mols of NaCl in it = mass of NaCl present / molar mass of NaCl

= 50 g / 58.5 g/mol

= 0.85 mols

So concentration of this solution = moles of solute / volume of solution in L

= 0.85 mols / 100 mL

= 0.854 mols / 0.1 L

= 8.54 M

= 8.54 * 1000 mM

= 8540 mM

That means this initial solution we have taken has a concentration (M1) = 8540 mM

Now to form a diluted solution with concentration (M2) = 100 mM, we can only add 1 part or 1% from this stock solution. i.e

volume from stock solution taken (V1) = 100 mL * 1% = 1 mL

Now we have to make a final concentration (M2) = 100 mM

So final volume of this solution (V2) = M1V1/M2 From the formula M1V1 = M2V2

= 8540 mM * 1 mL / 100 mM

= 85.4 mL

So here volume of 50% NaCl-solution taken = 1 mL

So volume of water added to dilute it to 85.4 mL = 85.4 mL - 1 = 84.4 mL

That means we have to use 84.4 mL or 84.4 parts of water.