According to the following equation- K2Cr2O7 - 4H2SO4- K2SO4 - Cr2SO43 - 4H2O - 3-O-the equivalent mass of K2Cr2O7 is

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Frankland Halide Reaction

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Question

According to the following equation,
$K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 4 H_{2} O + 3 [O]$
the equivalent mass of $K_{2} C r_{2} O_{7}$ is

m o l . m a s s / 3

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m o l . m a s s / 6

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m o l . m a s s

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m o l . m a s s / 12

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Solution

The correct option is C $m o l . m a s s / 3$
In the above reaction cromium changed its oxidation number from from +6 to +3 that is 3 electrons are gained.
So equivalent weight=molecular weight/number of electrons gained.
Therefore equivalent weight=molecular weight/3
Hence option A is correct.

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Similar questions

Q. The latent heatsof fusion in

J g^{- 1}

of five substances a (mol.mass =18); b(mol.mass =20); c(mol.mass =30), d(mol.mass =60) and e(mol.mass =30) are respectively

80, 45, 90, 45, 45

. Which of the following pair has same value of

Δ H_{f u s i o n}

Two half reactions for the following equation are:
$4 H_{2} S O_{4} + K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} ⟶ C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 N a_{2} S O_{4} + 4 H_{2} O$

Q. Determine the equivalent weight of

K_{2} C r_{2} O_{7}

and

F e S O_{4}

in the reaction.

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O

Q. Represent the following equation in ionic form.

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + 7 H_{2} O + K_{2} S O_{4}

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According to the following equation,K2Cr2O7+4H2SO4→K2SO4+Cr2(SO4)3+4H2O+3[O]the equivalent mass of K2Cr2O7 is

The correct option is C mol.mass/3In the above reaction cromium changed its oxidation number from from +6 to +3 that is 3 electrons are gained.So equivalent weight=molecular weight/number of electrons gained.Therefore equivalent weight=molecular weight/3Hence option A is correct.

According to the following equation,
$K_{2} C r_{2} O_{7} + 4 H_{2} S O_{4} \to K_{2} S O_{4} + C r_{2} (S O_{4})_{3} + 4 H_{2} O + 3 [O]$
the equivalent mass of $K_{2} C r_{2} O_{7}$ is

The correct option is C $m o l . m a s s / 3$
In the above reaction cromium changed its oxidation number from from +6 to +3 that is 3 electrons are gained.
So equivalent weight=molecular weight/number of electrons gained.
Therefore equivalent weight=molecular weight/3
Hence option A is correct.