You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VIII

Chemistry

Characteristics of Chemical Reaction

0.5 g of an o...

Question

$0.5 g$ of an oxalate was dissolved in water and the solution made to $100 m L$ . On titration $10 m L$ of this solution required $15 m L$ of $\frac{N}{20} K M n O_{4}$ . Calculate the percentage of oxalate in the sample.

Open in App

Solution

$15 m L \frac{N}{20} K M n O_{4} = 10 \times$ Normality of oxalate solution.
Normality of oxalate solution $= \frac{15}{20} \times \frac{1}{20} = \frac{3}{40}$
Strength of oxalate solution $= N o r m a l i t y \times$ Eq. mass of oxalate
$= \frac{3}{40} \times 44 = 3.3 g / L$ [Eq. mass of $C_{2} O_{4}^{2 -} = \frac{88}{2} = 44]$
Amount of oxalate in $100 m L$ solution $= \frac{3.3}{1000} \times 100 = 0.33 g$
% of oxalate $= \frac{0.33}{0.5} \times 100 = 66.0$

Suggest Corrections

Similar questions

0, 59

of an oxalate was dissolved in water and the solution made to

100 m l

. On tritation

10 m l

of this solution required

15 m l

\frac{N}{20} K M n O_{4}

. Calculate the percentage of oxalate in the sample.

1.17 g

of an impure sample of oxalic acid was dissolved and made up to

200 m L

with water.

10 m L

of this solution in acid medium required

8.5 m L

of a solution of potassium permanganate containing

3.16 g p e r l i t r e

of oxidation. Calculate the percentage purity of oxalic acid.

0.3 g

of an oxalate was dissolved in

100 m L

solution. The solution required

90 m L

\frac{N}{20} K M n O_{4}

for complete oxidation. The

%

of oxalate ion in salt is:

0.3 g of an oxalate salt was dissolved in 100 mL solution. The solution required 90 mLof $\frac{N}{20} K M n O_{4}$ for complete oxidation. The percentage of oxalate ion in the salt is:

Q. 2.0 g of chalk was dissolved in 10 ml of 4N HCl and the solution was made up to 100 ml. 25 ml of this solution required 18.75 ml of 0.2 N NaOH solution for complete neutralization. Calculate the percentage of pure

C a C O_{3}

in the sample of chalk?

Join BYJU'S Learning Program

0.5 g of an oxalate was dissolved in water and the solution made to 100 mL. On titration 10 mL of this solution required 15 mL of N20KMnO4. Calculate the percentage of oxalate in the sample.

15 mLN20KMnO4=10× Normality of oxalate solution.Normality of oxalate solution =1520×120=340Strength of oxalate solution =Normality× Eq. mass of oxalate=340×44=3.3 g/L [Eq. mass of C2O2−4=882=44]Amount of oxalate in 100 mL solution =3.31000×100=0.33 g% of oxalate =0.330.5×100=66.0

$0.5 g$ of an oxalate was dissolved in water and the solution made to $100 m L$ . On titration $10 m L$ of this solution required $15 m L$ of $\frac{N}{20} K M n O_{4}$ . Calculate the percentage of oxalate in the sample.