Question

If 0.290g of FAS is dissolved in 10mL of DI water and is titrated to equivalence with 12.23mL KMnO4, the molarity of KMnO4 is calculated to be: MnO4−(aq) + 5 Fe2+(aq) + 8 H+(aq) → Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(l)

Answer 1

Answer: Molarity of
`KMnO_4` is 0.121 M

Step-by-step explanation:

`MnO_4^-(aq)+5Fe^(2+)(aq)+8H+(aq)\rightarrow Mn^(2+)(aq)+5Fe^(3+)(aq)+4H_2O(l)`

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

`Molarity=(n* 1000)/(V_s)`

where,

n= moles of solute

`Moles=\frac{\text{Given mass}}{\text{Molar mass}}=(0.290g)/(392g/mol)=7.40* 10^(-4)mole`

`V_s` = volume of solution = 10 ml

`Molarity=(7.40* 10^(-4)* 1000)/(10)=0.074M`

According to the neutralization law:

`n_1M_1V_1=n_2M_2V_2`

where,

`M_1` = molarity of
`Fe^(2+)` solution = 0.074 M

`V_1` = volume of
`FAS` solution = 10 ml

`M_2` = molarity of
`KMnO_4` solution = ?

`V_2` = volume of
`KMnO_4` solution = 12.23 ml

`n_1` = valency of
`Fe^(2+)` = 2

`n_2` = valency of
`KMnO_4` = 1

`2* 0.074* 10=1* M_2* 12.23`

`M_2=0.121`

Therefore, the molarity of
`KMnO_4` is 0.121 M