Question

The mole fraction of glucose \[({C_6}{H_{12}}{O_6})\] in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer is?

Answer 1

Hint: Mole fraction is defined as the total number of the molecules that are present in the solution of a particular component that is divided by the total number of moles that are present in a mixture. It is a method by which we determine the total amount of concentration of a solution.

Formula used: Let us assume that the solution is made up of two components which are A and B, then mole fraction can be calculated by:
\[mole\,fraction\,of\,solution = \dfrac{{Moles\,of\,solute}}{{Moles\,of\,solute + Moles\,of\,solvent}}\]
\[ = \dfrac{{{n_A}}}{{{n_A} + {n_B}}}\]
\[Mole\,fraction\,of\,solvent = \dfrac{{moles\,of\,solvent}}{{moles\,of\,solute + moles\,of\,solvent}}\]
\[ = \dfrac{{{n_B}}}{{{n_A} + {n_B}}}\]

Complete Step by Step Solution:
To calculate the percentage of water in the solution, we need to calculate the total percentage of glucose that is present in the solution. It can be done as follows:
\[{X_{glu\cos e}} = 0.1\]
Mass percentage of glucose is equal to
\[ = \dfrac{{[0.1 \times 180]}}{{[0.1 \times 180 + 0.9 \times 180]}} \times 100\]
\[ = \dfrac{{1800}}{{[18 + 16.2]}}\]
\[ = [\dfrac{{1800}}{{34.2}}]\% \]
\[ = 52.63\% \]
\[ = 53\% \]
The percentage of glucose that is present is \[53\% \] , therefore the percentage of water in the glucose solution will be:
\[100\% - 53\% = \,\,47\% \]
Therefore, \[47\% \] of water is present in the 0.1m binary solution of \[({C_6}{H_{12}}{O_6})\] .

Note: The sum of mole fraction of all the components of a solution will always be equal to one. As the mole fraction represents the fraction of molecules and all the solutions will have different molecules with different masses, therefore the mole fraction will always be different from the mass fraction. Mole fraction is independent of the temperature of the solution and also independent of the individual densities of the solute and solvent in the solution.