If a 5.01 M aqueous solution of sodium hydroxide (NaOH, 40.00 g/mol) has a density of 1.21 g/mL, what is the mole fraction of sodium hydroxide?

Question

{hint: assume 1 L of solution} Enter your value using three significant figures.

Answer 1

1 L of solution x 1000 ml/L x 1.21 g/ml = 1210 g of solution.

5.01 mol/L NaOH x 40 g/mol = 200.4 g of NaOH

mass of water = 1210 g - 200.4 g = 1009.6 g

moles water = 1009.6 g x 1 mol/18 g = 56.1 moles water

Total moles = 5.01 + 56.1 = 61.11 = 61.1

mole fraction NaOH = 5.01/61.1 = 0.0820

Answer 2

Following the hint, assume a 1 L solution volume

Then, since it is 5.01 M NaOH, there are 5.01 moles of NaOH and 5.01 x 40 = 200.4 g NaOH

Let the number of moles of water be m_water.

The mass of water is then 14 m_water

The density is ( 200.4 + 14 m_water) / 1000 = 1.21

Solving for m_water gives m_water = 72.11 moles water

The mole fraction of NaOH is 5.01 /( 72.11 + 5.01) = .065