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Byju's Answer

Standard X

Chemistry

Mole Concept

Mass of KHC...

Question

Mass of $K H C_{2} O_{4}$ required to reduce 100 ml of 0.02 m $K M n O_{4}$ in acidic medium is x gram and to neutralise 100 ml of 0.05 m $C a (O H)_{2}$ is y gram then

x = y

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2 x = y

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x = 2 y

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none of these

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Solution

The correct option is B $2 x = y$

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Similar questions

Q. If x

g

is the mass of

N a H C_{2} O_{4}

is required to neutralise 100 mL of 0.2 (M)

N a O H

and y

g

is required to reduce 100 mL of 0.02 (M)

K M n O_{4}

in acidic medium, then:

Q. Mass of

K H C_{2} O_{4}

(potassium acid oxalate) required to reduce

100

mL of

0.02 M K M n O_{4}

in acidic medium (to

M n^{2 +})

x

g and to neutralise

100

mL of

0.05 M C a (O H)_{2}

y

g, then :

Q. If x

g

is the mass of

N a H C_{2} O_{4}

is required to neutralise 100 mL of 0.2 (M)

N a O H

and y

g

is required to reduce 100 mL of 0.02 (M)

K M n O_{4}

in acidic medium, then:

Q. Mass of

H_{2} C_{2} O_{4}

required to reduce 100 ml of 0.008M

K M n O_{4}

IN acidic medium is x g and to neutralize 100 ml of 0.01M

M g {(O H)}_{2}

IS y g, then the relation between x and y is

^{'} a^{'} g K H C_{2} O_{4}

required to reduce

100 m L

0.02 M K M n O_{4}

in acid medium and

^{'} b^{'} g K H C_{2} O_{4}

neutralises

100 m L

0.05 M C a (O H)_{2}

then:

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Mass of KHC2O4 required to reduce 100 ml of 0.02 m KMnO4 in acidic medium is x gram and to neutralise 100 ml of 0.05 m Ca(OH)2 is y gram then

The correct option is B 2x=y

Mass of $K H C_{2} O_{4}$ required to reduce 100 ml of 0.02 m $K M n O_{4}$ in acidic medium is x gram and to neutralise 100 ml of 0.05 m $C a (O H)_{2}$ is y gram then

The correct option is B $2 x = y$