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

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

Step-by-step 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 =
`(Given\;mass)/(Molar\;mass)`

Using the formula

Number of moles of NaOH
`(n_B)=(W_B)/(M_B)=(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)=(77.1)/(18.02)`

`n_A=4.279 moles`

Now, mole fraction of NaOH

=
`(n_B)/(n_B+n_A)`

`=(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