(C) The reaction for the deposition of magnesium is: $Mg^{2+} + 2e^- \rightarrow Mg(s)$.$2 \times 96500 \ C$ of charge is required to deposit $1 \text

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The reaction for the deposition of magnesium is: $Mg^{2+} + 2e^- \rightarrow Mg(s)$.$2 \times 96500 \ C$ of charge is required to deposit $1 \text{ mole}$ $(24 \ g)$ of $Mg$.Charge passed $= 9.65 \ C$.Moles of $Mg$ deposited $= \frac{9.65}{2 \times 96500} = \frac{1}{2 \times 10000} = 0.5 \times 10^{-4} = 5 \times 10^{-5} \text{ moles}$.Since $1 \text{ mole}$ of $Mg$ produces $1 \text{ mole}$ of Grignard reagent $(R-Mg-X)$,the moles of Grignard reagent obtained is $5 \times 10^{-5}$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

Medium

$9.65 \ C$ of electric current is passed through fused anhydrous $MgCl_2$. The magnesium metal thus obtained is completely converted into a Grignard reagent. The number of moles of Grignard reagent obtained is

KCET 2011 All KCET

A
$5 \times 10^{-4}$
B
$1 \times 10^{-4}$
C
$5 \times 10^{-5}$
D
$1 \times 10^{-5}$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Chemistry Questions

Chapter

Electrochemistry

Subtopic

Faraday’s law of electrolysis

Previous Year

KCET 2011 Paper All KCET years →

If in a voltaic cell $5 \ g$ of zinc is consumed,then how many ampere-hours are produced? (Given that the $E.C.E.$ of $Zn$ is $3.387 \times 10^{-7} \ kg/C$)

Medium

View Solution

$A$ current of strength $2.5 \ A$ was passed through $CuSO_4$ solution for $6 \ \text{minutes} \ 26 \ \text{seconds}$. The amount of copper deposited is .............. $g$ (Atomic weight of $Cu = 63.5$) $(1 \ F = 96500 \ C)$

Medium

View Solution

When $9.65 \ C$ of electricity is passed through a solution of silver nitrate (atomic weight of $Ag = 108$),the amount of silver deposited is $............ \ mg$.

Difficult

View Solution

$CuSO_4$ solution is electrolysed for $15 \ minutes$ to deposit $0.4725 \ g$ of copper at the cathode. The current in amperes required is (Faraday $= 96,500 \ C \ mol^{-1}$,atomic weight of copper $= 63$)

EasyAP EAMCET 2019

View Solution

On passing $96500 \ C$ of charge through a $CuSO_4$ solution,the amount of copper liberated is:

Easy

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$9.65 \ C$ of electric current is passed through fused anhydrous $MgCl_2$. The magnesium metal thus obtained is completely converted into a Grignard reagent. The number of moles of Grignard reagent obtained is

Similar Questions

If in a voltaic cell $5 \ g$ of zinc is consumed,then how many ampere-hours are produced? (Given that the $E.C.E.$ of $Zn$ is $3.387 \times 10^{-7} \ kg/C$)

$A$ current of strength $2.5 \ A$ was passed through $CuSO_4$ solution for $6 \ \text{minutes} \ 26 \ \text{seconds}$. The amount of copper deposited is .............. $g$ (Atomic weight of $Cu = 63.5$) $(1 \ F = 96500 \ C)$

When $9.65 \ C$ of electricity is passed through a solution of silver nitrate (atomic weight of $Ag = 108$),the amount of silver deposited is $............ \ mg$.

$CuSO_4$ solution is electrolysed for $15 \ minutes$ to deposit $0.4725 \ g$ of copper at the cathode. The current in amperes required is (Faraday $= 96,500 \ C \ mol^{-1}$,atomic weight of copper $= 63$)

On passing $96500 \ C$ of charge through a $CuSO_4$ solution,the amount of copper liberated is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module