$A$ solution of aluminium chloride is electrolysed for $30$ minutes using a current of $2 \ A$. The amount of the aluminium deposited at the cathode is $.................$
[Given : molar mass of aluminium and chlorine are $27 \ g \ mol^{-1}$ and $35.5 \ g \ mol^{-1}$ respectively,Faraday constant $= 96500 \ C \ mol^{-1}$]. (in $g$)
[Given : molar mass of aluminium and chlorine are $27 \ g \ mol^{-1}$ and $35.5 \ g \ mol^{-1}$ respectively,Faraday constant $= 96500 \ C \ mol^{-1}$]. (in $g$)
- A$0.336$
- B$1.007$
- C$1.660$
- D$0.441$
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Similar Questions
The number of electrons required to deposit $1 \ g$ atom of aluminium (at. wt. $= 27$) from a solution of aluminium chloride will be ............ $N$ (where $N$ is Avogadro's number).
MediumAIIMS 1992
View SolutionMatch List $I$ with List $II$.
List $I$ (Conversion) List $II$ (Number of Faraday required) $A$. $1 \text{ mole of } H_2O \text{ to } O_2$ $I$. $3F$ $B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$ $II$. $2F$ $C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$ $III$. $1F$ $D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$ $IV$. $5F$
Choose the correct answer from the options given below:
| List $I$ (Conversion) | List $II$ (Number of Faraday required) |
| $A$. $1 \text{ mole of } H_2O \text{ to } O_2$ | $I$. $3F$ |
| $B$. $1 \text{ mol of } MnO_4^- \text{ to } Mn^{2+}$ | $II$. $2F$ |
| $C$. $1.5 \text{ mol of } Ca \text{ from molten } CaCl_2$ | $III$. $1F$ |
| $D$. $1 \text{ mol of } FeO \text{ to } Fe_2O_3$ | $IV$. $5F$ |
Choose the correct answer from the options given below:
DifficultMHT CET 2026
View SolutionElectrolysis of $600 \ mL$ aqueous solution of $NaCl$ for $5 \ min$ changes the $pH$ of the solution to $12$. The current in Amperes used for the given electrolysis is $...........$. $(Nearest \ integer)$.
DifficultJEE MAIN 2025
View SolutionWhen $X$ amperes of current is passed through molten $AlCl_3$ for $96.5$ $s$,$0.09$ $g$ of aluminium is deposited. What is the value of $X$?
For the given reaction how much quantity of electricity in Coulomb is required? $\frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2$
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