Question

Mole fraction of the solute in a 1.00 molal aqueous solution is:
(a) 1.7700
(b) 0.1770
(c) 0.0177
(d) 0.0344

Answer 1

Molal term is used to express moles of the solute present in 1000 g of solvent. When the specific solvent is not given then water is taken as a solvent. Mole fraction of solute in the solution can be calculated as ${{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$
where ${{n}_{1}}$ is the number of moles of solvent and ${{n}_{2}}$ is the number of moles of the solute.

Complete answer:
Let us first understand about mole fraction and its calculation:
Mole fraction of a constituent (solute as well as solvent) is the fraction obtained by dividing the number of moles of that constituent by the total number of moles of all the constituents present in the solution. It is denoted by ‘x’.
If in a solution, ${{n}_{1}}$ and ${{n}_{2}}$ are the numbers of moles of solvent and the solute respectively, then
Mole fraction of solvent in the solution, ${{x}_{1}}=\dfrac{{{n}_{1}}}{{{n}_{1}}+{{n}_{2}}}$
Mole fraction of solute in the solution, ${{x}_{2}}=\dfrac{{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}$
The sum of mole fraction of all constituents of a solution is always equal to unity,
${{x}_{2}}+{{x}_{2}}+{{x}_{2}}+.....=1$
It may be noted that a mole fraction is a dimensionless quantity.
So the question says 1.00 molal aqueous solution, which means 1.00 moles of the solute is present in 1000 g of the solvent.
Hence the moles of solute = 1.00
Molar mass of the solvent (water) = 1000 g
Moles of solvent (water) = $\dfrac{1000}{\text{molar mass of water}}=\dfrac{1000}{18}=55.55$
Hence, the moles of solvent = 55.55
Now, for mole fraction of solute = $\dfrac{1}{1+55.55}=\dfrac{1}{56.55}=0.017683$
Hence the answer is 0.017683. This can be rounded off to 0.0177.
So, the correct answer is 0.0177.
So, the correct answer is “Option C”.

Note: Whenever you are calculating the mole fraction all the quantities must be converted into the number of moles. If the solvent is not mentioned you can directly apply the number of moles of the solvent 55.55.