Question

A solution contains $25%,25%$and $50%$of ${{H}_{2}}O,{{C}_{2}}{{H}_{5}}OH$and $C{{H}_{3}}COOH$by mass. The mole fraction of${{H}_{2}}O$would be
A.$0.25$
B.$2.5$
C.$0.503$
D.$5.03$

Answer 1

Hint: The mole fraction of any component can be determined by taking the ratio of the number of moles of a component in a solution to the total sum of the mole number of all the components present in that solution. For this first, we will have to calculate the number of moles of each molecule given in the question.

Formula Used:(i)Number of moles,$n=\dfrac{m}{M}$
Here $m\And M$ are the mass of a component and the molar mass of that component.
(ii) ${{X}_{1}}=\dfrac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}$ or,${{X}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$
${{X}_{1}}\And {{X}_{2}}$denotes the mole fraction of the component $1\And 2$.
${{n}_{1}}\And {{n}_{2}}$ represents the number of moles of the component $1\And 2$.

Complete answer:The given solution contains $25%$ ${{H}_{2}}O$ that means $25g$of ${{H}_{2}}O$is dissolved in $100g$of solution. Similarly, there are $25g$and $50g$of ${{C}_{2}}{{H}_{5}}OH$and $C{{H}_{3}}COOH$present in that solution of mass$100g$ .
The molar mass of ${{H}_{2}}O$, ${{M}_{{{H}_{2}}O}}$$=$($2\times $ Atomic weight of Hydrogen) + ($1\times $Atomic weight of oxygen)
$\therefore {{M}_{{{H}_{2}}O}}=(2\times 1)+(1\times 16)=18g/mol$
Now the number of moles of ${{H}_{2}}O$,${{n}_{{{H}_{2}}O}}$$=\dfrac{{{m}_{{{H}_{2}}O}}}{{{M}_{{{H}_{2}}O}}}=\dfrac{25g}{18g/mol}=1.4mol$
The molar mass of ${{C}_{2}}{{H}_{5}}OH$,${{M}_{{{C}_{2}}{{H}_{5}}OH}}=$ ($2\times $Atomic weight of carbon)$+$($6\times $Atomic weight of Hydrogen)$+$ ($1\times $Atomic weight of oxygen)
$\therefore {{M}_{{{C}_{2}}{{H}_{5}}OH}}=(2\times 12)+(6\times 1)+(1\times 16)=46g/mol$
Number of moles in ${{C}_{2}}{{H}_{5}}OH$,${{n}_{{{C}_{2}}{{H}_{5}}OH}}=$${{n}_{{{C}_{2}}{{H}_{5}}OH}}=\dfrac{{{m}_{{{C}_{2}}{{H}_{5}}OH}}}{{{M}_{{{C}_{2}}{{H}_{5}}OH}}}=\dfrac{25g}{46g/mol}=0.54mol$
The molar mass of $C{{H}_{3}}COOH$,${{M}_{C{{H}_{3}}COOH}}=$ ($2\times $Atomic weight of carbon)$+$($4\times $atomic weight of Hydrogen)$+$($2\times $Atomic weight of oxygen)
$\therefore {{M}_{C{{H}_{3}}COOH}}=(2\times 12)+(4\times 1)+(2\times 16)=60g/mol$
Number of moles in $C{{H}_{3}}COOH$,${{n}_{C{{H}_{3}}COOH}}=$ $=\dfrac{{{m}_{C{{H}_{3}}COOH}}}{{{M}_{C{{H}_{3}}COOH}}}=\dfrac{50g}{60g/mol}=0.8mol$
Mole fraction of ${{H}_{2}}O$,${{X}_{{{H}_{2}}O}}=\dfrac{{{n}_{{{H}_{2}}O}}}{{{n}_{{{H}_{2}}O}}+{{n}_{{{C}_{2}}{{H}_{5}}OH}}+{{n}_{C{{H}_{3}}COOH}}}$
Or,${{X}_{{{H}_{2}}O}}=\dfrac{1.4mol}{1.4mol+0.54mol+0.8mol}=0.51$
Therefore mole fraction of ${{H}_{2}}O$ would be 0.51

Thus, option (C) is correct.

Note: Generally mole fraction is a unitless quantity as it only gives the fraction of any component present in a particular solution. Also, it does not change with a temperature change. The temperature affects the volume of any solution but not the mass of any component.