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

Equivalent Mass

A 0.1097 g sa...

Question

A $0.1097 g$ sample of $A s_{2} O_{3}$ required $36.10 m L$ of $K M n O_{4}$ solution for its titration. The molarity of $K M n O_{4}$ solution is (Given Molecular mass of $A s_{2} O_{3} = 198$ )

0.02

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.04

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.012

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.03

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C 0.012
Let, molarity of $K M n O_{4}$ solution be M
$∴$ Eq. of $A s_{2} O_{3}$ = Eq. of $K M n O_{4}$ solution
$\frac{0.1097}{198} \times 4 = \frac{36.10 \times M \times 5}{1000}$ (Equivalent weight $A s_{2} O_{3} = \frac{198}{4}$ )
Molarity = 0.0122 M
Hence, (c) is the correct answer.

Suggest Corrections

Similar questions

Q. A

0.1097 g

sample of

A s_{2} O_{3}

required

36.10 m L

K M n O_{4}

solution for its titration. The molarity of

K M n O_{4}

solution is (Given Molecular mass of

A s_{2} O_{3} = 198

)

Q. A 0.1097 g sample of

A s_{2} O_{3}

required 26.10 mL of acidic

K M n O_{4}

solution for its titration. The molarity of

K M n O_{4}

solution is:

Q. A

0.46

g sample of

A s_{2} O_{3}

required

25.0

mL of

K M n O_{4}

solution for its titration. The molarity of

K M n O_{4}

solution is :

Q. A 3.4

g

sample of

H_{2} O_{2}

solution containing x% by mass requires

x

ml of a

K M n O_{4}

solution for complete oxidation under acidic medium. The molarity of

K M n O_{4}

solution is:

Q. A 3.4 g sample of

H_{2} O_{2}

solution containing

x % H_{2} O_{2}

by mass requires x mL of a

K M n O_{4}

solution for complete oixidation under acidic condition.The molarity of

K M n O_{4}

solution is:

Join BYJU'S Learning Program

Explore more

Equivalent Mass

Standard XI Chemistry

Join BYJU'S Learning Program

A 0.1097 g sample of As2O3 required 36.10 mL of KMnO4 solution for its titration. The molarity of KMnO4 solution is (Given Molecular mass of As2O3=198)

The correct option is C 0.012Let, molarity of KMnO4 solution be M ∴ Eq. of As2O3 = Eq. of KMnO4solution 0.1097198×4=36.10×M×51000 (Equivalent weight As2O3=1984) Molarity = 0.0122 M Hence, (c) is the correct answer.

A $0.1097 g$ sample of $A s_{2} O_{3}$ required $36.10 m L$ of $K M n O_{4}$ solution for its titration. The molarity of $K M n O_{4}$ solution is (Given Molecular mass of $A s_{2} O_{3} = 198$ )