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What is the mole fraction of Ba(OH)2 in an aqueous solution that contains 22.8% Ba(OH)2 by mass?

Respuesta :

Answer:

0.03

Explanation:

22.8 g Ba(OH)2 (1 mol Ba (OH)2/ 171.34 g) = 0.133 mol Ba (OH)2

77.2 g H2O (1 mol H2O/18 g) = 4.29 mol H2O

X= molar fraction= mol Ba(OH)2/ mol total

X= 0.133/ (0.133+4.29) = 0.03

0.03 is the mole fraction of [tex]Ba(OH)_2[/tex] in an aqueous solution that contains 22.8% [tex]Ba(OH)_2[/tex] by mass.

What is a mole fraction?

The ratio of the number of moles of one component of a solution or other mixture to the total number of moles representing all of the components.

Moles of [tex]Ba(OH)_2[/tex]

Moles = [tex]\frac{mass}{molar \;mass}[/tex]

Moles = [tex]\frac{22.8 g Ba(OH)_2}{171.34 g}[/tex]

= 0.133 mol [tex]Ba (OH)_2[/tex]

Moles of [tex]H_2O[/tex]

[tex]Moles = \frac{mass}{molar \;mass}[/tex]

[tex]Moles = \frac{77.2 g H_2O}{18 g}[/tex]

= 4.29 mol [tex]H_2O[/tex]

The mole fraction of [tex]Ba(OH)_2[/tex]

[tex]\frac{mol Ba(OH)_2}{total \;mol}[/tex]

= 0.03

0.03 is the mole fraction of [tex]Ba(OH)_2[/tex] in an aqueous solution that contains 22.8% [tex]Ba(OH)_2[/tex] by mass.

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