Home
Class 12
CHEMISTRY
Find out the % of oxalate ion in given s...

Find out the `%` of oxalate ion in given sample of oxalate salt of which 0.3 g is present in 100 mL of solution required 90 mL. `N//20KMnO_(4)` for complete oxidation.

Text Solution

AI Generated Solution

The correct Answer is:
To find the percentage of oxalate ion in the given sample of oxalate salt, we can follow these steps: ### Step 1: Determine the number of equivalents of KMnO4 used We know that the volume of KMnO4 solution used is 90 mL, and its normality is \( \frac{1}{20} \, N \). To calculate the number of equivalents of KMnO4: \[ \text{Number of equivalents} = \text{Volume (L)} \times \text{Normality} \] Convert 90 mL to liters: \[ 90 \, \text{mL} = 0.090 \, \text{L} \] Now, calculate the number of equivalents: \[ \text{Number of equivalents of KMnO4} = 0.090 \, \text{L} \times \frac{1}{20} \, N = 0.090 \times 0.05 = 0.0045 \, \text{equivalents} \] ### Step 2: Relate the equivalents of oxalate ion to KMnO4 Since the reaction is a complete oxidation, the number of equivalents of oxalate ion will be equal to the number of equivalents of KMnO4 used: \[ \text{Number of equivalents of oxalate ion} = 0.0045 \, \text{equivalents} \] ### Step 3: Calculate the number of moles of oxalate ion The n-factor for oxalate ion (C2O4^2-) is 2 because it can donate 2 electrons during oxidation. Thus, the number of moles of oxalate ion can be calculated as: \[ \text{Number of moles of oxalate ion} = \frac{\text{Number of equivalents}}{\text{n-factor}} = \frac{0.0045}{2} = 0.00225 \, \text{moles} \] ### Step 4: Calculate the mass of oxalate ion The molar mass of oxalate ion (C2O4^2-) is 88 g/mol. Therefore, the mass of oxalate ion can be calculated as: \[ \text{Mass of oxalate ion} = \text{Number of moles} \times \text{Molar mass} = 0.00225 \, \text{moles} \times 88 \, \text{g/mol} = 0.198 \, \text{g} \] ### Step 5: Calculate the percentage of oxalate ion in the sample The total mass of the oxalate salt sample is 0.3 g. The percentage of oxalate ion in the sample can be calculated as: \[ \text{Percentage of oxalate ion} = \left( \frac{\text{Mass of oxalate ion}}{\text{Mass of oxalate salt}} \right) \times 100 = \left( \frac{0.198 \, \text{g}}{0.3 \, \text{g}} \right) \times 100 \approx 66\% \] ### Final Answer: The percentage of oxalate ion in the given sample of oxalate salt is approximately **66%**. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

Calculate the percentage of oxalate ion in a given sample of oxalate salt of which 0.6 g dissolved 100 " mL of " water required 90 " mL of " (M)/(100)KMnO_(4) for complete oxidation.

Find out % of oxalate ion ina given sample of an alkali metal oxalate salt, 0.30g of it is dissolve in 100mL water and its required 90mL OF Centimolar KMnO_(4) solution in aicdic medium:

Find out % of oxalate ion in a given sample of an alkali metal oxalate salt, 0.30g of it is dissolved in 100mL water and its required 90mL of centimolar KMnO_4 solution in acidic medium :

25 g of a sample of FeSO_(4) was dissolved in water containing dil. H_(2)SO_(4) and the volume made upto 1 litre. 25 mL of this solution required 20mL of N/ 10 KMnO_(4) for complete oxidation. Calculate % of FeSO_(4).7H_(2)O in given sample.

10 g of oxalate was dissolved in 300 mL of solution. This solution required 250 mL of (M/10)KMnO_(4) in acidic medium for complete oxidation. The percentage purity of oxalate ion in the salt is :

3.92 g of ferrous ammonium sulphate crystals are dissolved in 100 ml of water, 20 ml of this solution requires 18 ml of KMnO_(4) during titration for complete oxidation. The weight of KMnO_(4) present in one litre of the solution is

0.6 g of a sample of pyrolusite was boiled with 200 " mL of " (N)/(10) oxalic acid and excess of dilute sulphuric acid. The liquid was filtered and the residue washed. The filtrate and washing were mixed and made up to 500 mL in a measuring flask. 100 " mL of " this solution required 50 " mL of " (N)/(30)KMnO_4 solution. Calculate the percentage of MnO_2 in the sample (Mn=55) .

A mixture of H_(2)SO_(4) and H_(2)C_(2)O_(4) (oxalic acid ) and some inert impurity weighing 3.185 g was dessolved in water and the solution made up to 1litre. 10 mL of this solution required 3 mL of 0.1 N NaOH for complete neutralization. In another experiment 100 mL of the same solution in hot condition required 4 mL of 0.02 M KMnO_(4) solution for complete reaction. The mass % of H_(2)SO_(4) in the mixture was:

A sample of MnSO_4.4H_2O is reacted in air to give Mn_3O_4 . The residue Mn_3O_4 is dissolved in 100 " mL of " (N)/(12)FeSO_4 containing H_2SO_4 The solution reacts completely with 50 " mL of " KMnO_4 . 25 " mL of " this KMnO_4 requires 30 " mL of " (N)/(10)FeSO_4 for complete oxidation determine the amount of MnSO_4.4H_2O in the sample.

A mixture of H_2SO_4 and H_2 C_2 O_4 (oxalic acid) and some inert impurity weighing 3.185g was dissolved in water and the solution made up to 1 litre. 10mL of this solution require 3mL of 0.1N Na OH for complete neutralization. In another experimant 100mL of the same solution in hot condition required 4mL of 0.02 MKMnO_4 solution for complete reaction. The mass % of H_2 SO_4 in the mixture was :