Question

The molality of 10% w /W NaOH

Answer 1

Let the mass of solution be x gram.

Then mass of NaOH present in it= x*10/100

Molar mass of NaOH= 40gram

So, number of moles of NaOh present in the solution = (10x/100)*1/40 = x/400

Now,mass of solvent present in the solution =mass of solution-mass of solvent = x-10x/100 =90x/100 (in grams)

So molality of solution=(x/400)/(90x/100)*1000 = 4.44