Question

A sodium hydroxide solution is made by mixing 8.70 g NaOH with 100 g of water. The resulting solution has a density of 1.087 g/mL.
What is the mass fraction and mole fraction of NaOH in the solution?

Answer 1

Answer:

Mass fraction = 0.08004

Mole fraction = 0.0377

Explanation:

Given, Mass of NaOH = 8.70 g

Mass of solution = 8.70 + 100 g = 108.70 g

$Mass\ fraction\ of\ NaOH=\frac {8.70}{108.70}$ = 0.08004

Molar mass of NaOH = 39.997 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{8.70\ g}{39.997\ g/mol}$

$Moles\ of\ NaOH= 0.2175\ mol$

Given, Mass of water = 100 g

Molar mass of water = 18.0153 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{100\ g}{18.0153\ g/mol}$

$Moles\ of\ water= 5.5508\ mol$

So, according to definition of mole fraction:

$Mole\ fraction\ of\ NaOH=\frac {n_{NaOH}}{n_{NaOH}+n_{water}}$

$Mole\ fraction\ of\ NaOH=\frac {0.2175}{0.2175+5.5508}=0.0377$