Question: Calculate the mole percentage of \[C{{H}_{3}}OH~\] and \[{{H}_{2}}O~\] respectively in \[60%\] (by m...

Question

Calculate the mole percentage of $C{{H}_{3}}OH~$ and ${{H}_{2}}O~$ respectively in $60%$ (by mass) aqueous Solution of $C{{H}_{3}}OH~$ .
A. $45.8,{ }54.2$
B. $54.2,{ }45.8$
C. $50,{ }50$
D. $60,{ }40$

Answer 1

Solution

Mole percentage of any component in a given sample is the amount of that component present in the total volume of the sample.
-After calculating the number of moles of each component present in the sample, the total number of moles in the sample could be calculated by simply adding the number of moles of both.
-Then we divide the individual number of moles with the total number of moles and multiply with hundred in order to get the mole percentage.

Complete step by step answer:
In order to calculate the mole percentage of the water and methanol present in the Solution, we will first calculate the number of moles of methanol and water individually, then we will add the total number of moles present in the Solution. Then we will calculate the mole percentage using the calculated values.
Given are the values of percentage by mass Solution of methanol.
$60%$ by mass Solution of methanol. This means in $100gm$ total Solution $60gm$ of methanol is dissolved in $40gm$ of water ${{H}_{2}}O~$
We know that the Molar mass of methanol is $32g/mol$
And the Molar mass of water is $18g/mol$
We know that the number of moles of solute is, given mass of solute, per molar mass of the Solution.
Which means No. of moles of $C{{H}_{3}}OH=\dfrac{60gm}{32g/mol}=1.875~mol~$
So the number of moles of methanol is $1.875mol$ and, we will calculate the number of moles of water,
And No. of moles of ${{H}_{2}}O=\dfrac{40gm}{18gm/mol}=2.222~mol~$
The number of moles of water is turned out to be $2.222mol$

Now we will calculate the total number of moles present in the Solution, that is
Total no. of moles $=1.875+2.222=4.097~moles~$
Now from this value we will calculate the mole percentage of both methanol and water separately, the mole percentage of a substance can be calculated by the following mathematical situation,
$\Rightarrow ~Mole{ }percentage=~\dfrac{Moles~of~compound}{Total~no.~of~moles}\times 100$
So we will calculate the mole percentage of methanol,
$Mole{ }%{ }of~C{{H}_{3}}OH=\dfrac{1.875}{4.097}\times 100{ }=45.8%~$
And similarly, now we will calculate the mole percentage of water
$Mole{ }%{ }of~H_2O=\dfrac{2.222}{4.097}\times 100{ }=54.23%~$
So the mole percentage of methanol is $45.8%$ and the mole percentage of water is turned out to be $54.23%$ . These values correspond to the values which are given in the option A.

So the correct option is option A.

Note: The percentage by mass ratio of methanol was given in the question, we used it to calculate the number of moles of methanol. And since the sample of mixture contained only two components, the remaining percentage of mass ratio would be of water.
-So we calculated the number of moles of water as well. Next we add up both the calculated number of moles to get the total number of moles in the sample.
-Then using this value, we calculate individual mole percentages of the component by simply dividing their number of moles with the total number of moles and multiplying it with hundred.