Question

A certain current liberates 0.504g of H in $2$ hours. How many grams of copper can be liberated by the same current flowing for the same time in $CuS{O_4}$, solution
A.$3.18g$
B.$16.0g$
C.$12.7g$
D.$63.5g$

Answer 1

Hint: Equivalent weight: It is defined as the ratio of weight of the substance to the n-factor.
The value of n-factor for base is defined as the number of hydroxide ions replaced by one mole of base. For example: the n-factor for sodium hydroxide is $1$. Similarly for acids the number of hydrogen ions replaced by one mole of acid, is known as n-factor of acid. For example: the n-factor of sulphuric acid is $2$.

Complete step by step solution:
We know that the mass liberated by copper sulphate solution will be equal to mass of hydrogen liberated.
We can calculate mass of liberated copper in copper sulphate solution as $Zit$ where $Z$ is the molar mass of the compound divided by its n-factor and $96500$ $i$ is currently passed and $t$ is time interval to which current is passed.
Here in the question we are given with the mass of hydrogen liberated in $2$ hours as $0.504g$.
By using the above formula i.e. mass of liberated hydrogen in copper sulphate solution as $Zit$, we can first find the value of current and then can use it to find the mass of liberated copper in copper sulphate solution. Or we can use the direct formula that the ratio of weight of hydrogen liberated to the equivalent weight of the hydrogen is equal to the ratio of weight of copper liberated to the equivalent weight of the copper because all other factors i.e. current and time are same so we can use this formula directly.
Equivalent weight: It is defined as the ratio of weight of the substance to the n-factor.
The value of n-factor for base is defined as the number of hydroxide ions replaced by one mole of base. For example: the n-factor for sodium hydroxide is $1$. Similarly for acids the number of hydrogen ions replaced by one mole of acid, is known as n-factor of acid. For example: the n-factor of sulphuric acid is $2$.
Now, the equivalent weight of hydrogen is $1$ and the equivalent weight of copper is $\dfrac{{63.5}}{2}$. So put these values in the formula: $\dfrac{{{\text{weight of hydrogen}}}}{{{\text{equivalent weight of hydrogen}}}}{\text{ = }}\dfrac{{{\text{weight of copper}}}}{{{\text{equivalent weight of copper}}}}$ put equivalent weight of copper as$\dfrac{{63.5}}{2}$, equivalent weight of hydrogen as $1$ and weight of hydrogen as
$0.504g$ we will get weight of copper as: ${\text{weight of copper = }}\dfrac{{63.5}}{2} \times 0.504 = 16.002g$.
Hence the weight of copper liberated is $16.0g$.

So, option B is correct.

Note: Primary batteries: Those batteries which cannot be rechargeable, are known as primary batteries. For example: dry cells.
Secondary batteries: Those batteries which can be rechargeable, are known as secondary batteries. For example: lead acid.