
A bottle labelled with “$12V\;{H_2}{O_2}$” contains $700\;mL$ solution. If a student mixes $300\;mL$ water in it. What is the g/litre strength & normality and volume strength of final solution?
Answer
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Hint: As we know that strength of solution is basically the volume in gram per litre, normality of a solution is defined as the gram equivalent of solute dissolved in $1000ml$ of solution and the volume strength of solution is the percentage of grams of solute in $100ml$ volume or volume strength is $5.6$ the normality of a solution.
Complete step by step solution: As we know that strength of solution is basically the volume in gram per litre and we are given a condition here that bottle is labelled with $12V\;{H_2}{O_2}$ which means that $1$ litre hydrogen peroxide gives $12$ litres of oxygen as a product. We also know that normality of a solution is defined as the gram equivalent of solute dissolved in $1000ml$ of solution or volume equivalent divided by $5.6$.
But the given volume of solution is $700mL$. We can first calculate the normality from this:
Normality of solution in $1L$or $1000mL$ will be given as the volume equivalents divided by $5.6$.
So we will get:
\[N = \dfrac{{Volume}}{{5.6}}\]
$\Rightarrow N = \dfrac{{12}}{{5.6}}$
$\Rightarrow N = 2.14N$
Now we can calculate the normality in $700mL$ will be:
$N = \dfrac{{2.14}}{{1000}} \times 700$
$\Rightarrow N = 1.5N$
Hence the normality of the solution is $1.5N$ as additional water will not change the normality of the solution.
Now we can calculate the volume strength of the solution which is the product of $5.6$ and normality.
$Volume\;strength = 5.6 \times N$
$\Rightarrow Vol.\;strength = 5.6 \times 1.5 = 8.4g{L^{ - 1}}$
Hence volume strength is $5.6 \times 1.5 = 8.4g{L^{ - 1}}$
Note: Always remember that the Normality is used to measure the concentration of a solution and can be expressed in terms of equivalent per litre where $1$ Normal solution can be said to contain one litre or volume equivalent of solute in a thousand millilitre of solution. And volume strength is a unique term for hydrogen peroxide and it generally means that $1$ volume strength of hydrogen peroxide will liberate $1$ volume strength of oxygen.
Complete step by step solution: As we know that strength of solution is basically the volume in gram per litre and we are given a condition here that bottle is labelled with $12V\;{H_2}{O_2}$ which means that $1$ litre hydrogen peroxide gives $12$ litres of oxygen as a product. We also know that normality of a solution is defined as the gram equivalent of solute dissolved in $1000ml$ of solution or volume equivalent divided by $5.6$.
But the given volume of solution is $700mL$. We can first calculate the normality from this:
Normality of solution in $1L$or $1000mL$ will be given as the volume equivalents divided by $5.6$.
So we will get:
\[N = \dfrac{{Volume}}{{5.6}}\]
$\Rightarrow N = \dfrac{{12}}{{5.6}}$
$\Rightarrow N = 2.14N$
Now we can calculate the normality in $700mL$ will be:
$N = \dfrac{{2.14}}{{1000}} \times 700$
$\Rightarrow N = 1.5N$
Hence the normality of the solution is $1.5N$ as additional water will not change the normality of the solution.
Now we can calculate the volume strength of the solution which is the product of $5.6$ and normality.
$Volume\;strength = 5.6 \times N$
$\Rightarrow Vol.\;strength = 5.6 \times 1.5 = 8.4g{L^{ - 1}}$
Hence volume strength is $5.6 \times 1.5 = 8.4g{L^{ - 1}}$
Note: Always remember that the Normality is used to measure the concentration of a solution and can be expressed in terms of equivalent per litre where $1$ Normal solution can be said to contain one litre or volume equivalent of solute in a thousand millilitre of solution. And volume strength is a unique term for hydrogen peroxide and it generally means that $1$ volume strength of hydrogen peroxide will liberate $1$ volume strength of oxygen.
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