
Find the concentration of ${H_2}S{O_4}$ in the battery after $6.0A$ of current drawn from the battery for $13.40h$ if a fully charged battery contains $500ml$ of $5M$ ${H_2}S{O_4}$.
Answer
577.2k+ views
Hint:Convert the time given into seconds by multiplying it with 3600 and using the relation between charge, current and time find the amount of charge. Concentration of the substance is the quantity of solute present in the given quantity of solution.
Complete step by step answer:
Concentration is the abundance of constituents divided by the total volume of mixture. The concentration can be any kind of chemical mixture but mostly the solutes and solvents in the solution.
Let the current be $i$ and time be $t$. Therefore, according to the question, the given quantities are –
$
i = 6A \\
t = 13.4h = 13.4 \times 60 \times 60 \\
\Rightarrow t = 48240\sec \\
$
Now, we have to find the amount of charge flowing. So, from the definition of current, we know that, current is the amount of charge flowing per second. This can be represented mathematically as –
$i = \dfrac{q}{t}$
where, $q$ is the amount of charge
So, the above relation can also be written as –
$q = i \times t$
Therefore, putting the values of current and time in the above equation
$
\Rightarrow q = 6 \times 48240 \\
\therefore q = 289440C \\
$
So, the amount of charge flown is equal to the 3 moles of electron.
The dissociation reaction of ${H_2}S{O_4}$ is –
From the above reaction, we can see that 1.5 moles of ${H_2}S{O_4}$ is dissociated.
Therefore, the initial amount of ${H_2}S{O_4}$ can be calculated as –
$ \Rightarrow \dfrac{1}{2} \times 5 = 2.5mol$
Final amount of ${H_2}S{O_4}$ is 1 mol.
So, the concentration of ${H_2}S{O_4}$ is calculated by –
$
\Rightarrow \dfrac{{1mol}}{{500}} \times 100m \\
\Rightarrow 2M \\
$
Hence, the concentration of ${H_2}S{O_4}$ in the battery is $2M$.
Note:
Concentration can be described in the qualitative way. To make the solution concentrated we have to reduce the amount of solute or add more solute.
Make sure that the dissociation reaction of ${H_2}S{O_4}$ is balanced, so that we can get the number of moles dissociated.
Complete step by step answer:
Concentration is the abundance of constituents divided by the total volume of mixture. The concentration can be any kind of chemical mixture but mostly the solutes and solvents in the solution.
Let the current be $i$ and time be $t$. Therefore, according to the question, the given quantities are –
$
i = 6A \\
t = 13.4h = 13.4 \times 60 \times 60 \\
\Rightarrow t = 48240\sec \\
$
Now, we have to find the amount of charge flowing. So, from the definition of current, we know that, current is the amount of charge flowing per second. This can be represented mathematically as –
$i = \dfrac{q}{t}$
where, $q$ is the amount of charge
So, the above relation can also be written as –
$q = i \times t$
Therefore, putting the values of current and time in the above equation
$
\Rightarrow q = 6 \times 48240 \\
\therefore q = 289440C \\
$
So, the amount of charge flown is equal to the 3 moles of electron.
The dissociation reaction of ${H_2}S{O_4}$ is –
From the above reaction, we can see that 1.5 moles of ${H_2}S{O_4}$ is dissociated.
Therefore, the initial amount of ${H_2}S{O_4}$ can be calculated as –
$ \Rightarrow \dfrac{1}{2} \times 5 = 2.5mol$
Final amount of ${H_2}S{O_4}$ is 1 mol.
So, the concentration of ${H_2}S{O_4}$ is calculated by –
$
\Rightarrow \dfrac{{1mol}}{{500}} \times 100m \\
\Rightarrow 2M \\
$
Hence, the concentration of ${H_2}S{O_4}$ in the battery is $2M$.
Note:
Concentration can be described in the qualitative way. To make the solution concentrated we have to reduce the amount of solute or add more solute.
Make sure that the dissociation reaction of ${H_2}S{O_4}$ is balanced, so that we can get the number of moles dissociated.
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