Question

When 6 gm urea dissolves in 180 gm ${{H}_{2}}O$. The mole fraction of urea is:-
(A) $\dfrac{10}{10.1}$
(B) $\dfrac{10.1}{10}$
(C) $\dfrac{10.1}{0.1}$
(D) $\dfrac{0.1}{10.1}$

Answer 1

Hint: As we know that mole fraction is one of the ways to express the concentration of a solution or a mixture. It is the number of moles of a single component of a mixture divided by the total number of moles of the mixture or solution. So here we have to calculate the mole fraction of urea in a mixture of water and urea itself.
Formula used: We will use the following formulas:-
${{x}_{a}}=\dfrac{{{n}_{a}}}{\text{Total moles}}$
Where,
${{x}_{a}}$ = mole fraction of component ‘a’
$n=\dfrac{w}{M}$
where,
n = number of moles
w = mass of the element or compound
M = molar mass of the compound

Complete answer:
First we will calculate the moles of all the components of the given solution i.e., urea and water followed by the calculation of mole fraction of urea as follows:-
-Calculation of moles of urea ($N{{H}_{2}}CON{{H}_{2}}$):-
Atomic mass of N = 14 g/mol
Atomic mass of C = 12 g/mol
Atomic mass of O = 16 g/mol
Atomic mass of H = 1 g/mol
Molar mass of urea ($N{{H}_{2}}CON{{H}_{2}}$) = (2(14) + 12 + 16 + 4(1)) g/mol = 60 g/mol
The given value of mass of urea is (w) = 6 grams
On substituting all the values in the formula of number of moles, we get:-
$\begin{align}
  & \Rightarrow n=\dfrac{w}{M} \\
 & \Rightarrow n=\dfrac{6g}{60g/mol} \\
 & \Rightarrow n=0.1mole \\
\end{align}$
-Calculation of moles of water (${{H}_{2}}O$):-
Atomic mass of O = 16 g/mol
Atomic mass of H = 1 g/mol
Molar mass of water (${{H}_{2}}O$) = (16 + 2(1)) g/mol = 18 g/mol
The given value of mass of ${{H}_{2}}O$ is (w) = 180 grams
On substituting all the values in the formula of number of moles, we get:-
$\begin{align}
  & \Rightarrow n=\dfrac{w}{M} \\
 & \Rightarrow n=\dfrac{180g}{18g/mol} \\
 & \Rightarrow n=10moles \\
\end{align}$
-Calculation of mole fraction of urea:-
$\begin{align}
  & \Rightarrow {{x}_{urea}}=\dfrac{{{n}_{urea}}}{\text{Total moles}} \\
 & \Rightarrow {{x}_{urea}}=\dfrac{{{n}_{urea}}}{{{n}_{urea}}+{{n}_{{{H}_{2}}O}}} \\
 & \Rightarrow {{x}_{urea}}=\dfrac{0.1mole}{0.1mole+10moles} \\
 & \Rightarrow {{x}_{urea}}=\dfrac{0.1}{10} \\
\end{align}$

Therefore the mole fraction of urea is (D) $\dfrac{0.1}{10.1}$

Note:
-Always remember to check the options so as to calculate the answer in respective forms. If the options are provided in fractions then we do not require any further calculation.
-Also mole fraction is a unitless quantity as units cancel each other and kindly use values along with units in the formulas.