Question

$0,59$ of an oxalate was dissolved in water and the solution made to $100ml$. On tritation $10ml$ of this solution required $15ml$ of $\frac{N}{20}KMn{O}_4$. Calculate the percentage of oxalate in the sample.

• A. $66.0$
• B. $663$
• C. $0.66$
• D. $33.33$

Answer 1

Answer: (A)

$66.0$

$0.5g$ oxalate (impure) in $100ml$

$10ml$ of that solution needed $15ml$ of $\frac{N}{20}{KMnO}_4$

$N_1{V}_1=N_2\times V_2$

$\left(\frac{N}{20}\right)\times \frac{15}{1000}=\left(2\right)\times \left(M\right)\times \frac{10}{100}$

$M=\frac{1\times 15}{20\times 20}\Rightarrow \frac{3}{80}$

Oxalate mol $=\left(\frac{0.5}{90}\right)\times \frac{1000}{100}=\frac{1}{18}$

$x$% of $\frac{1}{18}$ is $\frac{3}{18}$ ($\because$ $x$% pure assumed)

$\frac{x}{100}\times \frac{1}{18}=\frac{3}{80}$

$x\approx 66$%