Question

10% w/w NaOH Aqueous solution’s value of Molality ?

Answer 1

Let the mass of solution be x gram.
Then mass of NaOH present in it= x*20/100
Molar mass of NaOH= 40gram
So, number of moles of NaOh present in the solution = (20x/100)*1/40 = x/200
Now,mass of solvent present in the solution =mass of solution-mass of solvent = x-20x/100 =80x/100 (in grams)
So molality of solution=(x/200)/(80x/100)*1000 =6.25