Question

A 10.0m solution of NaOH has a density of 1.33 g/cm3 at 20 °C. Calculate the mole fraction, weight percent (%), and molarity (M) of NaOH?

Mole fraction =

Weight percent = %

Molarity = M

Answer 1

1.

10.0 m solution of NaOH means,

10 mol of NaOH is in 1 kg of water.

So, moles of NaOH = 10.0 mol

Moles of water = 1000 / 18 = 55.6 mol

Mole fraction of NaOH = moles of NaoH / Total moles of solution

x = 10.0 / (10.0 + 55.6)

x = 0.152

2.

Mass of NaOH = moles * molar mass

Mass of NaOH = 10.0 * 40.0 = 400. g.

Mass percentage NaOH = Mass of NaOH * 100 / Total mass of solution

= 400. * 100 / (1400)

= 28.6 %

3.

Total mass of solution = 1400 g.

Given that, density d = 1.33 g/ mL

SO, volume of 1.33 g. of solution = 1 mL

then, volume of 1400 g. of solution = 1400 * 1 / 1.33 = 1077. mL = 1.08 L

Therefore,

Molarity = moles of solute / volume of solution in L

M = 10.0 / 1.08

M = 9.26 M