Question

Can anyone solve this? From a \[95\% \] ethanol solution, prepare a \[125ml\] of a \[75\% \] ethanol solution.

Answer 1

Hint: According to the law of conservation of mass the number of moles in both the ethanol solutions are the same. Thus, the volume of the ethanol needed to prepare a \[125ml\] of a \[75\% \] ethanol solution can be calculated from the below formula. From that volume the preparation can be known.
Formula used:
\[{M_1}{V_1} = {M_2}{V_2}\]
\[{M_1}\] is initial molarity
\[{V_1}\] is the initial volume. let it be \[x\]
\[{M_2}\] is final molarity
\[{V_2}\] is final volume

Complete answer:
Ethanol is an alcohol with the molecular formula of \[{C_2}{H_5}OH\] it has two carbon atoms, six hydrogen atoms and one oxygen atom. It has one oxygen-hydrogen bond which can be called as a most polar bond, Due to the presence of this bond these are known as hydroxyl compounds.
Given percent of ethanol solution is \[95\% \] the molarity will be \[0.95M\]
The final molarity will be \[0.75M\]
The final volume is \[125ml\]
According to the law of conservation of mass the number of moles in both the ethanol solutions are the same. By applying the above formula,
\[0.95 \times x = 0.75 \times 125\]
The value of \[x\] will be \[98.684ml\]
Thus, \[98.684ml\] was needed to make \[75\% \] ethanol solution. But the final volume is \[125ml\]
So, the remaining amount will be water which is \[125 - 98.684 = 26.316ml\]
To prepare \[125ml\] of \[75\% \] ethanol solution \[98.684ml\] of ethanol and \[26.316ml\] of water must be added.

Note:
Given are the percentage of solutions, to convert into molarity they should divide with a factor of \[100\] . In this way both the molarities were obtained. In both the solutions percent was different, but prepared by mixing the same amount of mass and molar mass. Thus, the number of moles is also constant.