Question

Calculate the mole fraction of the total ions in an aqueous solution prepared by dissolving 0.400 moles of MgBr2 in 850.0 g of water

Answer 1

 moles water = 850.0 g / 18.02 g/mol=47.2 
moles Mg2+ = 0.400 
moles Cl- = 2 x 0.400 = 0.800 
moles ions = 0.400 + 0.800= 1.2 
mole fraction ions = 1.2 / 1.2 + 47.2 =0.0248

Answer 2

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.

$\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar mass of water}}=\frac{850.0g}{18g/mole}=47.22moles$

Now we have to calculate the moles of magnesium ion and bromide ion.

As we are given that 0.4 moles of $MgBr_2$ 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.

$\chi_{\text{(Total ions)}}=\frac{n_{\text{(Total ions)}}}{n_{\text{(Total ions)}}+n_{water}}$

$\text{Mole fraction of total ions in an aqueous solution}=\frac{1.2mole}{1.2mole+47.22mole}=0.0248$

Therefore, the mole fraction of the total ions in an aqueous solution is 0.0248