Question

Pure ethanol is difficult to prepare and therefore expensive, \[95\% \] (v/v) ethanol is much cheaper. Consequently, \[95\% \] ethanol is generally used in the preparation of dilute water solutions of ethanol. It has been found that \[70\% \] (v/v) ethanol concentration is the most effective for preoperative skin disinfection. How much \[95\% \] ethanol would be needed to produce a \[2.5{\text{ }}L\]solution of \[70\% \] ethanol?

Answer 1

Hint: \[95\% \] (v/v) of ethanol means \[95ml\]of ethanol diluted in $100ml$of water. Same goes for $70\% $(v/v) ethanol, $70ml$of ethanol diluted in $100ml$of water. Using this, calculate the amount required. It is given (v/v) ratio so ethanol will be taken in milliliters and not in gm.

Complete answer:
We need to find how much of \[95\% \] ethanol is required to prepare \[2.5{\text{ }}L\]solution of \[70\% \] ethanol
We know that \[70\% \]of ethanol means \[70ml\] of ethanol diluted in $100ml$of water
This can be written as:
$100ml\xrightarrow{{}}70ml$
$ \Rightarrow 1ml\xrightarrow{{}}\dfrac{{70}}{{100}}ml$
We need \[2.5{\text{ }}L\]of solution and so we will convert milliliters to liters.
So, $ \Rightarrow 1{\text{ L}}\xrightarrow{{}}\dfrac{{70}}{{100}} \times 1000ml$ (as$1L = 1000ml$)
$ \Rightarrow 2.5{\text{ L}}\xrightarrow{{}}\dfrac{{70}}{{100}} \times 1000 \times 2.5ml$
$ \Rightarrow 70 \times 25ml$
Thus, by this we found out that \[2.5{\text{ }}L\]of solution contains $70 \times 25ml$of \[70\% \]ethanol
Now we need to calculate the quantity of \[95\% \](v/v) ethanol that is required for preparing \[2.5{\text{ }}L\]solution of \[70\% \] ethanol. We will calculate it in the same manner we did for \[70\% \] ethanol.
\[95\% \]Of ethanol means \[95ml\]of ethanol diluted in $100ml$of water.
So $95ml\xrightarrow{{}}100ml$
$ \Rightarrow 1ml\xrightarrow{{}}\dfrac{{100}}{{95}}ml$
$ \Rightarrow 70 \times 25ml\xrightarrow{{}}\dfrac{{100}}{{95}} \times 70 \times 25ml$
\[ \Rightarrow 1842.10ml\]
Therefore \[1842.10ml\]of \[95\% \] ethanol is required for the preparation of \[2.5{\text{ }}L\]solution of \[70\% \] ethanol. This means that \[1.84210{\text{ L}}\]of \[95\% \] ethanol is required to prepare \[2.5{\text{ }}L\]solution of \[70\% \] ethanol.

Note:
\[70\% \]or \[95\% \] of ethanol is given in (v/v) i.e. volume by volume ratio. Sometimes it might be given (w/v) this is weight by volume ratio. This means that in case of \[70\% \]of ethanol, \[70gm\]of ethanol is diluted in \[100ml\]of water. And for \[95\% \] of ethanol \[95gm\]of ethanol diluted in \[100ml\]of water.