The mole fraction of glucose ($$C_6H_{12}O_6$$) in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is .........

We are told that the mole fraction of glucose, represented as $$X_{\,\text{glucose}}$$, is $$0.1$$ in an aqueous binary solution. A binary solution contains only two components, here glucose ($$C_6H_{12}O_6$$) and water ($$H_2O$$). By definition of mole fraction we have

$$X_{\,\text{glucose}}=\dfrac{n_{\,\text{glucose}}}{n_{\,\text{glucose}}+n_{\,\text{water}}}=0.1$$

where $$n_{\,\text{glucose}}$$ and $$n_{\,\text{water}}$$ denote the number of moles of glucose and water, respectively.

Because the total of all mole fractions in a binary mixture is $$1$$, the mole fraction of water must satisfy

$$X_{\,\text{glucose}}+X_{\,\text{water}}=1\;\;\Longrightarrow\;\;0.1+X_{\,\text{water}}=1$$

Hence

$$X_{\,\text{water}}=1-0.1=0.9$$

To handle the numerical work conveniently, we choose a simple basis: let the total number of moles in the solution be $$1$$ mole. Under this basis

$$n_{\,\text{glucose}}=X_{\,\text{glucose}}\times1\text{ mol}=0.1\text{ mol}$$

$$n_{\,\text{water}}=X_{\,\text{water}}\times1\text{ mol}=0.9\text{ mol}$$

Now we convert these amounts to masses using molar masses. The molar mass of glucose ($$C_6H_{12}O_6$$) is $$180\;\text{g mol}^{-1}$$, and that of water ($$H_2O$$) is $$18\;\text{g mol}^{-1}$$.

Mass of glucose:

$$m_{\,\text{glucose}}=n_{\,\text{glucose}}\times M_{\,\text{glucose}}=0.1\;\text{mol}\times180\;\text{g mol}^{-1}=18\;\text{g}$$

Mass of water:

$$m_{\,\text{water}}=n_{\,\text{water}}\times M_{\,\text{water}}=0.9\;\text{mol}\times18\;\text{g mol}^{-1}=16.2\;\text{g}$$

The total mass of the solution is therefore

$$m_{\text{total}}=m_{\,\text{glucose}}+m_{\,\text{water}}=18\;\text{g}+16.2\;\text{g}=34.2\;\text{g}$$

We now need the mass percentage of water. By definition,

$$\text{Mass \% of water}=\dfrac{m_{\,\text{water}}}{m_{\text{total}}}\times100$$

Substituting the values obtained:

$$\text{Mass \% of water}=\dfrac{16.2\;\text{g}}{34.2\;\text{g}}\times100 =\left(0.473684\dots\right)\times100 =47.3684\dots$$

Rounding to the nearest integer gives

$$47$$

So, the answer is $$47$$.