
$3.65gm$of $HCl$dissolved in$16.2$ $gms$of water. The mole fraction of $HCl$in the resulting solution is [EAMCET$2003$]
A.$0.4$
B.$0.3$
C.$0.2$
D.$0.1$
Answer
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Hint: Generally mole fraction is the ratio of the number of moles of one component of a particular solution to the total number of moles of all the components present in that solution. To calculate the mole fraction of $HCl$, we will need to take the ratio of the number of moles $HCl$to the sum of the number of moles of $HCl$ water(${{H}_{2}}O$).
Formula Used:No. of moles,$n=\dfrac{m}{M}$
$m=$mass of component
$M=$ molar mass of the component
Mole fraction of $HCl$,${{X}_{HCl}}=\dfrac{{{n}_{HCl}}}{{{n}_{HCl}}+{{n}_{{{H}_{2}}O}}}$
Here ${{n}_{HCl}}=$ number of moles of $HCl$
${{n}_{{{H}_{2}}O}}=$Number of moles of ${{H}_{2}}O$
Complete answer:In this problem $HCl$is dissolved in water forming an aqueous $HCl$solution. First, we have to calculate the number of moles of components present in the aqueous $HCl$solution.
The molar mass of HCl,${{M}_{HCl}}$ = Atomic weight of $H$$+$ Atomic weight of $Cl$
$\therefore {{M}_{HCl}}=(1+35.5)gm/mol$ $=36.5gm/mol$
Given, the mass of$HCl$, ${{m}_{HCl}}=3.65gm$
$\therefore
{{n}_{Hcl}}=\dfrac{{{m}_{HCl}}}{{{M}_{HCl}}}=\dfrac{3.65gm}{36.5gm/mol}=0.1mol$
The molar mass of ${{H}_{2}}O$,${{M}_{H2O}}=$( $2\times $ Atomic weight of $H$)$+$Atomic weight of $O$
${{M}_{H2O}}=(2\times 1+16)gm/mol=18gm/mol$
Given the mass of water,${{m}_{{{H}_{2}}O}}=16.2gm$
$\therefore {{n}_{{{H}_{2}}O}}=\dfrac{{{m}_{{{H}_{2}}O}}}{{{M}_{{{H}_{2}}O}}}=\dfrac{16.2gm}{18gm/mol}=0.90mol$
Now we can calculate the mole fraction of $HCl$ by taking the ratio of the number of moles of $HCl$to the total sum of the number of moles of $HCl$and ${{H}_{2}}O$.
Thus, the mole fraction of $HCl$,${{X}_{HCl}}=$$\dfrac{{{n}_{HCl}}}{{{n}_{{{H}_{2}}O}}+{{n}_{HCl}}}$
Or,${{X}_{HCl}}=\dfrac{0.1mol}{0.90mol+0.1mol}=0.1$
Hence, the mole fraction of $HCl$in the resulting solution is $0.1$.
Thus, option (D) is correct.
Note: Mole fraction is a unitless quantity as it expresses a fraction of components and the total sum of mole fractions of all the components present in the solution is always equal to one. Also, it is temperature independent as it prevents variations whenever temperature fluctuates in the system. This is because temperature only changes volume but not mass.
Formula Used:No. of moles,$n=\dfrac{m}{M}$
$m=$mass of component
$M=$ molar mass of the component
Mole fraction of $HCl$,${{X}_{HCl}}=\dfrac{{{n}_{HCl}}}{{{n}_{HCl}}+{{n}_{{{H}_{2}}O}}}$
Here ${{n}_{HCl}}=$ number of moles of $HCl$
${{n}_{{{H}_{2}}O}}=$Number of moles of ${{H}_{2}}O$
Complete answer:In this problem $HCl$is dissolved in water forming an aqueous $HCl$solution. First, we have to calculate the number of moles of components present in the aqueous $HCl$solution.
The molar mass of HCl,${{M}_{HCl}}$ = Atomic weight of $H$$+$ Atomic weight of $Cl$
$\therefore {{M}_{HCl}}=(1+35.5)gm/mol$ $=36.5gm/mol$
Given, the mass of$HCl$, ${{m}_{HCl}}=3.65gm$
$\therefore
{{n}_{Hcl}}=\dfrac{{{m}_{HCl}}}{{{M}_{HCl}}}=\dfrac{3.65gm}{36.5gm/mol}=0.1mol$
The molar mass of ${{H}_{2}}O$,${{M}_{H2O}}=$( $2\times $ Atomic weight of $H$)$+$Atomic weight of $O$
${{M}_{H2O}}=(2\times 1+16)gm/mol=18gm/mol$
Given the mass of water,${{m}_{{{H}_{2}}O}}=16.2gm$
$\therefore {{n}_{{{H}_{2}}O}}=\dfrac{{{m}_{{{H}_{2}}O}}}{{{M}_{{{H}_{2}}O}}}=\dfrac{16.2gm}{18gm/mol}=0.90mol$
Now we can calculate the mole fraction of $HCl$ by taking the ratio of the number of moles of $HCl$to the total sum of the number of moles of $HCl$and ${{H}_{2}}O$.
Thus, the mole fraction of $HCl$,${{X}_{HCl}}=$$\dfrac{{{n}_{HCl}}}{{{n}_{{{H}_{2}}O}}+{{n}_{HCl}}}$
Or,${{X}_{HCl}}=\dfrac{0.1mol}{0.90mol+0.1mol}=0.1$
Hence, the mole fraction of $HCl$in the resulting solution is $0.1$.
Thus, option (D) is correct.
Note: Mole fraction is a unitless quantity as it expresses a fraction of components and the total sum of mole fractions of all the components present in the solution is always equal to one. Also, it is temperature independent as it prevents variations whenever temperature fluctuates in the system. This is because temperature only changes volume but not mass.
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