Molarity of 50% (w/w) Sodium Hydroxide (NaOH)

OVERVIEW

50 grams of solute in 100 grams of solution

  • A 50% (w/w) concentrated Sodium hydroxide solution is a clear colorless liquid. It can be purchased from many suppliers.
  • A 50% (w/w) sodium hydroxide solution means that there is 50 g of NaOH per 100 g of solution. 
  • The density of 50% (w/w) Sodium hydroxide solution is 1.515 g/ml at 25°C which means that the weight of the 1 ml of Sodium hydroxide solution is 1.515 g at 25°C.
  • Molarity refers to the number of moles of the solute present in 1 liter of solution.
  • In simple words, 1 mole is equal to the atomic weight of the substance. For example, 1 mole of Sodium hydroxide is equal to 40.00 grams of Sodium hydroxide (NaOH, molecular weight = 40.00).
  • To calculate the molarity, one must first calculate how much Sodium hydroxide is present in 1 L of 50% Sodium hydroxide solution. Once we know the amount of sodium hydroxide present in 1 L solution, we can calculate the molarity of the solution by dividing the NaOH amount by the molecular weight.

50% sodium hydroxide to moles per liter

Calculation procedure:

Known values 
Molar mass of Sodium Hydroxide
40.00 g/mole
Concentration of Sodium Hydroxide solution50% (% by mass, w/w)
Density of 50% (w/w) Sodium Hydroxide solution
1.515 g/ml

Step 1: Calculate the volume of 100 grams of Sodium Hydroxide solution.
Formula:

Density = WeightVolume 

OR

Volume = WeightDensity 

The volume of 100 g of Sodium hydroxide solution: 001.515  = 66.0066 ml

Note: 50% (w/w) Sodium Hydroxide means that 100 g of solution contains 50 g of Sodium Hydroxide.

The volume of 100 g of Sodium hydroxide solution is 66.0066 ml which means 50 g of NaOH is present in 66.0066 ml of  Sodium hydroxide solution.

Step 2: Calculate how many grams of NaOH is present in 1000 ml of Sodium hydroxide solution.
66.0066 ml of Sodium Hydroxide solution contains   = 50 grams of NaOH
1 ml of Sodium hydroxide solution will contain      =   5066.0066 grams of NaOH 

1000 ml of Sodium hydroxide solution will contain = 1000 x 5066.0066    = 757.50 grams of NaOH

1000 ml of Sodium hydroxide solution will contain 757.50 grams of NaOH.

Step 3: Calculate the number of moles of NaOH present in 757.50 grams of NaOH.
40 g of NaOH is equal to 1 mole.

1 g of NaOH will be equal to 140  moles.

757.50 g of NaOH will be equal to = 757.50 x 140 = 18.9375 moles 


Therefore, we can say that 1 liter of sodium hydroxide solution contains 18.9375 moles or in other words molarity of 50% (w/w) Sodium Hydroxide is equal to 18.9375 M.

Flow diagram: Molarity calculation of 50% sodium hydroxide solution

Calculator – Calculate the molarity of concentrated Sodium Hydroxide (NaOH)
Use Calculator to calculate the molarity of concentrated Sodium Hydroxide (NaOH) when concentration is given in % by mass (w/w)

Sodium Hydroxide (NaOH) Molecular weight: 40.00 g/mole


Concentration of Sodium Hydroxide in % (wt/wt) : % (wt/wt)

(Change the % (wt/wt) concentration)


Density of glacial Sodium Hydroxide: g/ml

(Change the density)



Molarity of Sodium Hydroxide: 18.938 M


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4 thoughts on “Molarity of 50% (w/w) Sodium Hydroxide (NaOH)

  1. 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

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