SATHEE: Redox Reactions And Electrochemistry Ques 719

Redox Reactions And Electrochemistry Ques 719

Question: On electrolyzing the molten chloride of metal $725 m l$ of $C l_{2}$ liberated at $25^{\circ} C$ and 740 mm Hg pressure at anode for every one gm of metal deposite at cathode. The formula of metal chloride is (GAM of metal $= 52.01$ )

Options:

A) $M C I$

B) $M C I_{2}$

C) $M C I_{3}$

D) $M C I_{4}$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] equivalents of metal produced = equivalents of $C l_{2}$ gas produced $M C l_{x}$ $x_{M} = x_{C l_{2}}$ $x \times n_{M} = 2 \times n_{C l_{2}}$ $x \times \frac{1}{52.01} = 2 \times \frac{740}{760} \times \frac{725 \times 10^{- 3}}{0.0821 \times 298}$

$\Rightarrow$ $x = 3$ $M C l_{3}$

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Redox Reactions And Electrochemistry Ques 719

Question: On electrolyzing the molten chloride of metal 725ml of Cl2 liberated at 25∘C and 740 mm Hg pressure at anode for every one gm of metal deposite at cathode. The formula of metal chloride is (GAM of metal =52.01 )

Options:

Answer:

Solution:

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Question: On electrolyzing the molten chloride of metal $725 m l$ of $C l_{2}$ liberated at $25^{\circ} C$ and 740 mm Hg pressure at anode for every one gm of metal deposite at cathode. The formula of metal chloride is (GAM of metal $= 52.01$ )