Calculate the molality of 10 % (w/w) aqueous caustic sode. What is the mole fraction of water in the solution?

Explanation

100 grams solution = 10 grams of NaOH + 90 grams of water
Molality (m) `= (10)/(40) xx 1000/90 = 100/36 = 2.78 mol g^(-1)`
Number of moles of `NaOH = n_1 = 10//40 = 0.25`
Number of moles of `H_2O = n_2 = 90//18 = 5.0`
Mole - fraction of water in the solution = `n_2//(n_1 + n_2) = (5.0)/(5.25) = 0.9523`.