Question

Find out the percentage purity of impure sample of Oxalic acid. You are
supplied M/100 KMnO4 solution.

​

Answer 1

$\bold{Percentage\,ofpurity\,of\,oxalic\,acid=\frac{Strength\,of\,pure\,sample}{Strength\,of\,given\,sample}\times 100}$

Step by step explanation:

The balanced chemical equation of potassium permanganate is as follows.

$\bold{2MnO_{4}^{-}+16H^{+}+5C_{2}O_{4}^{2-}\rightarrow 2Mn^{2+}+8H_{2}O+10CO_{2}}$

From the above balanced chemical equation of potassium permanganate,

It is clear that 5 moles of oxalic acidd treated with two moles of permanaganate.

Hence, the molarity of $KMnO_{4}$ is as follows.

$\frac{M_{KMnO_{4}}\times V_{KMnO_{4}}}{M_{oxalic\,acid}\times V_{oxalic\,acid}}= \frac{2}{5}$.......................(1)

Molarity of potassium permanganate = 1/00

Volume of oxalic acid = 20.

Substitute the values in equation (1)

$\frac{\frac{1}{100}\times V_{KMnO_{4}}}{M_{oxalic\,acid}\times 20}= \frac{2}{5}$

• $\bold{M_{oxalic\,acid}=\frac{x}{100}}$

Let's calculate the strength of oxalic acid.

• $\bold{Strength\,of\,oxalic\,acid=Molarity\times Molarmass}$

$=\frac{x}{800}\times 126 y(g/l)$

• $\bold{Percentage\,ofpurity\,of\,oxalic\,acid=\frac{Strength\,of\,pure\,sample}{Strength\,of\,given\,sample}\times 100}$

$=\frac{y}{4}\times 100$

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