Question

What is mole fraction of fe 2+ in aqueous solution if it's concentration is 0.1 m

Answer 1

The mole fraction of $Fe^{2+}$ ions in the aqueous solution is 0.0018

Explanation:

We are given:

Molality of $Fe^{2+}$ ions = 0.1 m

This means that 0.1 moles of $Fe^{2+}$ ions are present in 1 kg or 1000 g of water

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of water = 1000 g

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$\text{Moles of water}=\frac{1000g}{18g/mol}=55.56mol$

Total moles of solution = [0.1 + 55.56] = 55.66 moles

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

• For $Fe^{2+}$ ions:

$\chi_{Fe^{2+}}=\frac{n_{Fe^{2+}}}{n_{Fe^{2+}}+n_{H_2O}}\\\\\chi_{Fe^{2+}}=\frac{0.1}{55.66}=0.0018$

• For water:

$\chi_{H_2O}=\frac{n_{H_2O}}{n_{Fe^{2+}}+n_{H_2O}}\\\\\chi_{H_2O}=\frac{55.55}{55.66}=0.9982$

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