What is the molarity (M) of 50% NaOH?

I know the formula mass Na = 22.9, O = 16, H = 1and density of 50% NaOH = 1.53

formula mass for NaOH = 40 approximately

and I write

X mol NaOH = 50 g NaOH (1 mol NaOH/ 40 gNaOH)
= 1.25 moles of NaOH

100g NaOH soln. (1ml NaOH soln./1.53 NaOH soln.)
= 65. 36 ml = 0.0654 L

M = # of moles of solute/ # of liters of solution
= 1.25 moles NaOH/0.0654 NaOH soln
= 0.019 M NaOH

This is not correct, can someone tell me where I have gone wrong?

Well, from what I see, you divided it wrong. because if you divide anything by a decimal, it has to get bigger. You already converted the units, so 1.25 mol / 0.0654L = 19.125M.

Another way to do it (the way I would have done it), is to assume there is 1L of the solution. (take 1L as a basis). So in 1L of solution, there is 1.53g/ml * 1000mL = 1530g of NaOH solution. Half of this is NaOH, so 765g

this 765g has to be converted to mol's by dividing by the molar mass, which is 40g/mol, so 765g/40g/mol = 19.125 mols in 1L.

So 19.125M.

This should be the right answer.
Hope that helped
M = # of moles of solute/ # of liters of solution
= 1.25 moles NaOH/0.0654 L NaOH soln
= 0.019 M NaOH ->>> wrong

= 1.25 moles NaOH/0.0654 L NaOH soln
= 19.125 M NaOH
I do not know why there is so much of confusion?

50% NaOH=50 gms per 100 ml.=500 grams per 1000ml(litre).
Molarity=molecular weight in grams(gram molecule=mole) per litre.molecular weight of NaOH=40

500/40=12.5 Molar
A typical % soln. is w/v- weight per volume, so it's 50g/100mL
Which is 500g NaOH/L
now multiply by (1mol NaOH/40g NaOH)
which leaves
12.5 mol NaOH/L
so it's 12.5M

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