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Equivalent Mass

30 mL of a so...

Question

$30 m L$ of a solution containing $9.15 g L^{- 1}$ of an oxalate $K_{x} H_{y} (C_{2} O_{4})_{z} . n H_{2} O$ are required for titrating $27 m L$ of $0.12 N N a O H$ and $36 m L$ of $0.12 M K M n O_{4}$ separately.
The value of $x + y + z$ is:

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Solution

Let, molecular weight of salt is $M$
(i) n-factor in acid-base titration = $y$
(ii) n-factor in redox titration $= 2 \times z$
$meq of acid in 30 mL = meq of NaOH$
$30 \times \frac{9.15}{M} \times y = 27 \times 0.12 30 \times \frac{9.15}{M} \times (2 \times z) = 36 \times 0.12 ∴ \frac{y}{z} = \frac{3}{2}$
Total cationic charge = total anionic charge
$\Rightarrow x + y = 2 z ∴ x : y : z = 1 : 3 : 2$

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Similar questions

30 m L

of a solution containing

9.15 g m L^{- 1}

K_{x} H_{y} (C_{2} O_{4})_{z} . n H_{2} O

are required for titrating

27 m L

0.12 N N a O H

36 m L

0.12 M K M n O_{4}

separately.
The value of

x + y + z

30

mL of a solution containing

9.15

g/L of an oxalate,

K_{x} H_{y} (C_{2} O_{4})_{z} . n H_{2} O

is required for titration

27

0.12 N N a O H

36

0.12 N K M n O_{4}

Calculate the value of

x

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9.15 g L^{- 1}

K_{X} Y_{X} (C_{2} O_{4})_{z} . n H_{2} O

required for titration

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36 m l

0.12 N K M n O_{4}

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10 m l

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H C l

25 m l

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H_{2} {S O}_{4}

40 m l

X

N

H {N O}_{3}

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100 m l

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Equivalent Mass

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