Question

3.92g of mohrsalt(mol wt=392) is present in 100 ml of an aqueous solution.the sulphate ion concentration in molarity of the resulting solution?

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Answer 1

The concentration of sulfate ion in the solution is 0.2 M

Explanation:

The chemical formula of Mohr salt is $(NH_4)_2Fe(SO_4)_2.6H_2O$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Given mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Given mass of Mohr salt = 3.92 g

Molar mass of Mohr salt = 392 g/mol

Volume of solution = 100 mL

Putting values in above equation, we get:

$\text{Molarity of Mohr salt}=\frac{3.92\times 1000}{392\times 100}=0.1M$

The chemical equation for the ionization of mohr salt follows:

$(NH_4)_2Fe(SO_4)_2.6H_2O\rightarrow 2NH_4^++Fe^{2+}+2SO_4^{2-}+6H_2O$

1 mole of mohr salt produces 2 moles of ammonium ions, 1 mole of iron ions, 2 moles of sulfate ions and 6 moles of water

So, concentration of sulfate ions = (2 × 0.1) = 0.2 M

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Answer 2

Explanation:

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