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Sagot :
Answer:
Following are the solution to the given points:
Explanation:
Oxalic acid volume [tex]= 25.00 \ mL = 0.0250 \ litres[/tex]
KMnO4 volume [tex]= 12.70 \ ml = 0.0127 \ litres[/tex]
KMnO4 molarity [tex]= 0.0206\ M = 0.0206 \ \frac{mol}{l}[/tex]
In point a:
Its pink presence after full intake of oxalic acid with attachment to KMnO4 is suggested by the end-point of the process due to the small abundance of KMnO4, As just a self predictor, KMnO4 is used.
In point b:
[tex]H_2C_2O_4[/tex] molecules mole ratio to [tex]MnO_4^-[/tex] ions:
The equilibrium for both the oxalic acid and KMnO4 reaction is suggested:
[tex]6H+ (aq) + 2MnO_4- (aq) + 5H_2C_2O_4 (aq) \rightarrow 10CO_2 (g) + 8H_2O (l) + 2Mn_2+ (aq)[/tex]
The reaction of 5 mol of oxalic acid is 2 mol [tex]MnO_4^-[/tex] ions
[tex]H_2C_2O_4[/tex]: molecules mole proportion to [tex]MnO_4^-[/tex] ions:
[tex]5 H_2C_2O_4[/tex]: : [tex]2MnO_4^-[/tex]
In point c:
The Moles of [tex]MnO_4^-[/tex] ions reacted with the [tex]H_2C_2O_4[/tex]:
The molar mass of the solution is the number of solute moles in each volume of water
[tex]Molarity =\frac{moles}{Volume}\\\\Moles \ of\ KMnO_4 = Molarity \times volume[/tex]
Moles with ions reacted to mol with both the amount of : supplied.
In point d:
[tex]H_2C_2O_4[/tex] moles in the sample present:
[tex]H_2C_2O_4[/tex] moles = moles [tex]MnO_4^-[/tex] ions [tex]\times[/tex] mole ratio
[tex]H_2C_2O_4[/tex] moles in the sample = [tex]2.6162 \times 10^{-4}\ mol \times (\frac{5}{2})[/tex]
[tex]H_2C_2O_4[/tex] molecules = [tex]6,5405\times 10^{-4}[/tex] mol are present in the sample
In point e:
Oxalic acid molarity = [tex]\frac{mole}{volume}[/tex]
[tex]=\frac{ 6.54 \times 10^{-4} mol}{0.025\ L} \\\\ = 0.0260 \ M[/tex]
In point f:
Oxalic acid level by mass in the solution:
Oxalic acid mass calculation:
Oxalic acid molar weight = 90.0349 [tex]\frac{g}{mol}[/tex].
Oxalic acid mass per liter = oxalic acid moles per liter [tex]\times[/tex] molar mass
[tex]= 0.0260 \frac{mol}{L} \times 90.0349 \frac{g}{mol}\\\\= 2.3409 \frac{g}{L}\\\\ = 2.3409 \frac{g}{1000 \ mL}\\\\= 0.2409 \frac{g}{100 \ mL}[/tex]
When Oxalic acid solution density[tex]= 1.00 \ \frac{g}{mL}[/tex]
Mass oxalic acid percentage = [tex]0.2409 \%[/tex]
Oxalic acid mass proportion [tex]= 0.24\% \ \frac{W}{v} \ \ Mass[/tex]
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