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

$95\%$ (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.