Question

What is the mole fraction of NaOH in an aqueous solution that
contains 22.9% NaOH by mass?
Na=23
O=16
H= 1.01

Answer 1

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles =$\frac{Given\;mass}{Molar\;mass}$

Using the formula

Number of moles of  NaOH$(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}$

$n_B=0.572moles$

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water$(n_A)=\frac{77.1}{18.02}$

$n_A=4.279 moles$

Now, mole fraction of NaOH

=$\frac{n_B}{n_B+n_A}$

$=\frac{4.279}{0.572+4.279}$

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882