Comments on: Molarity of 50% (w/w) Sodium Hydroxide (NaOH)

By: Topic: Sodium Hydroxide (NaOH) - Laboratory Notes

Topic: Sodium Hydroxide (NaOH) - Laboratory Notes — Sat, 17 Dec 2022 07:33:09 +0000

[…] Molarity of 50% (w/w) Sodium Hydroxide (NaOH) […]

By: Andrew

Andrew — Thu, 03 Feb 2022 00:00:34 +0000

40 g NaOH solid in 100 mL solution is 10% w/v. 40g/(40 g/mol) = 1 mol / 0.1 L = 10%.

By: Preparation of 10 M Sodium Hydroxide from 50% (w/w) Stock Solution - Laboratory Notes

Sun, 24 Oct 2021 07:30:36 +0000

[…] Step 1: Calculate the molarity of 50% (w/w) concentrated Sodium Hydroxide solutionThe procedure to calculate molarity of 50% (w/w) concentrated Sodium Hydroxide solution is described in detail in the post “Molarity of 50% (w/w) Sodium Hydroxide (NaOH)”. […]

By: Paul

Paul — Thu, 28 Jan 2021 13:40:37 +0000

Ok so you lost me on a couple of specifics. I get all the calculations. However, it seems to me when it’s stated “ Calculate the volume of 100 grams of Sodium Hydroxide solution”, the solution concentration has to be assumed at the 50% w/w. And by converting it to a volume, to me at least, you are now calculating for a 50% v/v solution. I, mistakenly, would have said a 50% w/w solution would be 50g NaOH in 50ml (50g) of water. So 498.6g NaOH + 498.6g H₂O (H₂O = 0.9973g/ml @75℉, 1atm). The molarity is 498.6/40=12.46 moles in…(ok, now I get it). I was going to say, in 997.3 ml of solution which would be 12.5 molar. But you need the density to determine total volume, or grams/liter. Never mind. And thanks