Answer : The mole fraction of the total ions in an aqueous solution is 0.0248
Explanation :
First we have to calculate the moles of water.
[tex]\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{850.0g}{18g/mole}=47.22moles[/tex]
Now we have to calculate the moles of magnesium ion and bromide ion.
As we are given that 0.4 moles of [tex]MgBr_2[/tex] that means, there are 0.4 moles of magnesium ion and 0.8 moles of bromide ions.
Total mole of ions in an aqueous solution = 0.4 + 0.8 = 1.2 mole
Now we have to calculate the mole fraction of the total ions in an aqueous solution.
[tex]\chi_{\text{(Total ions)}}=\frac{n_{\text{(Total ions)}}}{n_{\text{(Total ions)}}+n_{water}}[/tex]
[tex]\text{Mole fraction of total ions in an aqueous solution}=\frac{1.2mole}{1.2mole+47.22mole}=0.0248[/tex]
Therefore, the mole fraction of the total ions in an aqueous solution is 0.0248
