Question

Calcular las fracciones molares de los componentes de una disolución formada por: 250 g de agua, 50 g de ácido acético y 25 g de cloruro de sodio.

Answer 1

The question is: Calculate the molar fractions of the components of a solution made up of: 250 g of water, 50 g of acetic acid and 25 g of sodium chloride.

Answer: The mole fraction of water, acetic acid and sodium chloride is 0.916, 0.054, and 0.028.

Explanation:

As number of moles is the mass of substance divided by its molar mass.

Therefore, moles of given species are calculated as follows.

The number of moles of water (molar mass = 18 g/mol):

$Moles = \frac{mass}{molar mass}\\= \frac{250 g}{18 g/mol}\\= 13.88 mol$

The number of moles of acetic acid (molar mass = 60.052 g/mol):

$Moles = \frac{mass}{molar mass}\\= \frac{50 g}{60.052 g/mol}\\= 0.832 mol$

The number of moles of sodium chloride (molar mass = 58.44 g/mol):

$Moles = \frac{mass}{molar mass}\\= \frac{25 g}{58.44 g/mol}\\= 0.427 mol$

Therefore, mole fraction of each component is calculated as follows.

Mole fraction of water:

$Mole fraction = \frac{moles of water}{Total moles}\\= \frac{13.88}{13.88 + 0.832 + 0.427}\\= \frac{13.88}{15.139}\\= 0.916$

Mole fraction of acetic acid:

$Mole fraction = \frac{moles of water}{Total moles}\\= \frac{0.832}{13.88 + 0.832 + 0.427}\\= \frac{0.832}{15.139}\\= 0.054$

Mole fraction of NaCl:

$Mole fraction = \frac{moles of water}{Total moles}\\= \frac{0.427}{13.88 + 0.832 + 0.427}\\= \frac{0.427}{15.139}\\= 0.028$

Thus, we can conclude that mole fraction of water, acetic acid and sodium chloride is 0.916, 0.054, and 0.028.