An experiment is performed to standardize a methyl alcohol solution by titration
ID: 713394 • Letter: A
Question
An experiment is performed to standardize a methyl alcohol solution by titration with potassium permanganate according to the unbalanced reaction below:CH3OH(aq) + KMnO4(aq) ==> MnO2(s) + HCO2H
If 100 ml of aqueous solution with methanol is titrated by 38.40 ml of 0.0492M potassium permanganate solution, what is the concentration of methanol in g/100ml of solution? An experiment is performed to standardize a methyl alcohol solution by titration with potassium permanganate according to the unbalanced reaction below:
CH3OH(aq) + KMnO4(aq) ==> MnO2(s) + HCO2H
If 100 ml of aqueous solution with methanol is titrated by 38.40 ml of 0.0492M potassium permanganate solution, what is the concentration of methanol in g/100ml of solution?
CH3OH(aq) + KMnO4(aq) ==> MnO2(s) + HCO2H
If 100 ml of aqueous solution with methanol is titrated by 38.40 ml of 0.0492M potassium permanganate solution, what is the concentration of methanol in g/100ml of solution? CH3OH(aq) + KMnO4(aq) ==> MnO2(s) + HCO2H
If 100 ml of aqueous solution with methanol is titrated by 38.40 ml of 0.0492M potassium permanganate solution, what is the concentration of methanol in g/100ml of solution?
Explanation / Answer
Ans. Moles of KMnO4 consumed = Molarity x Volume of solution in liters
= 0.0492 M x 0.03840 L
= 0.00188928 mol
# Balanced reaction: 2 KMNO4 + 3 CH3OH ---> 3 HCOH + 2 KOH + 2 H2O + 2 MNO2
According to the stoichiometry of balanced reaction, 2 mol KMnO4 neutralizes 3 mol CH3OH.
So,
Moles of CH3OH in sample = (3/2) x Moles of KMnO4 consumed
= (3/2) x 0.00188928 mol
= 0.00283392 mol
# Now,
Mass of methanol = Moles x Molar mass
= 0.00283392 mol x (32.04216 g/ mol)
= 0.090805 g
Therefore, [methanol] = 0.090805 g / 100.0 mL
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