Question

# What is the mole fraction, X, of solute and the molality, m (or b), for an...

What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 19.0% NaOH by mass?

hint: Molality, m (or b), is the number of moles of solute per kilogram of solvent. Mole fraction, X, is the number of moles of a component divided by the total number of moles. Choose an arbitrary mass of solution, such as 100 g, and determine the masses of NaOH and H2O. Then convert those values to moles or kilograms as necessary.

19 % NaOH by mass that mean in 100 gm of solution contain 19 gm of NaOH and 81 gm of water

molar mass of NaOH = 39.997 gm/mole then 19 gm NaOH = 19/39.997 = 0.475 mole

molar mass of water = 18.01528 gm/mole then 81 gm of water = 81/18.01528 = 4.496 mole

total mole = 0.475 + 4.496 = 4.971 mole

mole fraction = no. of mole of indivisual molecule / total mole of solution

mole frction of NaOH = 0.475 / 4.971 = 0.09555

mole fraction of NaOH(X) =  0.09555

molality = no.of mole of solute / solvent in kilogram

no. of mole of NaOH = 0.475 mole

100 gm = 0.1 Kg

molality of NaOH = 0.475 / 0.1 = 4.75 m

molality of NaOH = 4.75 m