# A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution?

Question:

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL−1, then what shall be the molarity of the solution?

Solution:

10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 − 10) g = 90 g of water.

Molar mass of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol−1

Then, number of moles of glucose $=\frac{10}{180} \mathrm{~mol}$

= 0.056 mol

$\therefore$ Molality of solution $=\frac{0.056 \mathrm{~mol}}{0.09 \mathrm{~kg}}=0.62 \mathrm{~m}$

Number of moles of water $=\frac{90 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}$

= 5 mol

$\Rightarrow$ Mole fraction of glucose \begin{aligned}\left(x_{\mathrm{g}}\right) &=\frac{0.056}{0.056+5} \\ &=0.011 \end{aligned}

And, mole fraction of water $x_{\mathrm{w}}=1-x_{\mathrm{g}}$

= 1 − 0.011

= 0.989

If the density of the solution is 1.2 g mL−1, then the volume of the 100 g solution can be given as:

$=\frac{100 \mathrm{~g}}{1.2 \mathrm{~g} \mathrm{~mL}^{-1}}$

$=83.33 \mathrm{~mL}$

$=83.33 \times 10^{-3} \mathrm{~L}$

$\therefore$ Molarity of the solution $=\frac{0.056 \mathrm{~mol}}{83.33 \times 10^{-3} \mathrm{~L}}$

= 0.67 M