Question

The weight of kmno4 that can oxidise 100ml of 0.2M oxalic acid in acidic medium is

Answer 1

Answer:

1.2643 g

Explanation:

Considering:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For Oxalic acid :

Molarity = 0.2 M

Volume = 100 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 100 × 10⁻³ L  = 0.1 L

Thus, moles of Oxalic acid :

$Moles=0.2 \times {0.1}\ moles$

Moles of Oxalic acid  = 0.02 moles

The reaction of $KMnO_4$ and Oxalic acid is shown below as;

$2KMnO_4+3H_2SO_4+5(COOH)_2\rightarrow K_2SO_4+2MnSO_4+8H_2O+10CO_2$

5 moles of oxalic acid reacts with 2 moles of $KMnO_4$

1 mole of oxalic acid reacts with 2/5 moles of $KMnO_4$

0.02 moles of oxalic acid reacts with (2/5)*0.02 moles of $KMnO_4$

Moles of $KMnO_4$ = 0.008 moles

Molar mass of $KMnO_4$ =  158.034 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$0.008\ moles= \frac{Mass}{158.034\ g/mol}$

Mass of $KMnO_4$ = 1.2643 g