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.

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Solution

Mole Fraction:-
Mole fraction of the solute is defined as the moles of the solute with respect to the total moles.
$M o l e f r a c t i o n = \frac{M o l e s o f s o l u t e}{T o t a l m o l e s}$
Mass percentage:-
percentage of the mass of the solute with respect to the mass of the solution.
$M o l e p e r c e n t a g e = \frac{M a s s o f s o l u t e}{T o t a l m a s s o f t h e s o l u t i o n} \times 100$
Let the total number of the moles is $1$
Then mole fraction of the glucose is given $0.1$ and mole fraction of the water will be $0.9$ .
$The number of the moles of the solute = \frac{w e i g h t o f t h e s o l u t e}{m o l e c u l a r w e i g h t}$
$\begin{array}{rcl} Weight of the solute & = & Number of moles \times molecular weight \\ molecular weight of the glucose & = & 180 \\ = & 0.1 \times 180 \\ = & 18 \end{array}$
Similarly,
$\begin{array}{rcl} Weight of the solvent & = & Number of moles \times molecular weight \\ Molecular weight of the water & = & 18 \\ = & 0.9 \times 18 \\ = & 16.2 \end{array}$
Now, Apply the formula of the Mass percentage
$\begin{array}{rcl} Mole percentage of water & = & \frac{Mass of water}{Total mass of the solution} \times 100 \\ = & \frac{16.2}{(16.2 + 18)} \times 100 \\ = & \frac{16.2}{34.2} \times 100 \\ = & 47.36 \\ = & 47 % \end{array}$
Hence, The mass percentage of water in it, to the nearest integer, is $47 %$ .

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