Question

A solution of glucose in water is labelled as \(10\%\) \(\text{w/w}\), what would be the molality and mole fraction of each component in the solution? If the density of solution is \(1.2\, \text{g mL}^{–1}\), then what shall be the molarity of the solution?

Answer 1

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

Molar mass of glucose \((\text{C}_6\text{H}_{12}\text{O}_6) = 6 \times12 + 12 \times 1 +6 \times16 = 180\, \text{g mol}^{-1}\)

Then, number of moles of glucose\(=\frac{10}{180}\text{mol}\)
\(= 0.056 \text{mol}\)

∴Molality of solution\(= \frac{0.056\, \text{mol}}{0.09\, \text{kg}} = 0.62 \text{m}\)

Number of moles of water \(= \frac{90\text{g}}{18\,\text{gmol}^{-1}}\)
\(= 5 \text{mol}\)

⇒Mole fraction of glucose\((x_g) = \frac{0.056}{(0.056+5)}\)
\(=0.011\)

And, mole fraction of water \(\text{X}_\text{w} = 1- \text{X}_\text{g}\)
\(= 1-0.011\)
\(= 0.989\)
If the density of the solution is \(1.2\, \text{g mL}^{-1}\), then the volume of the 100g solution can be given as:

\(\frac{100g }{1.2\text{gmL}^{-1}}\)

\(= 83.33 \text{mL}\)

\(= 83.33\times10^{-3} L\)

∴Molarity of the solution\(= \frac{0.056\text{mol}}{83.33\times10^{-3} \text{L}}\)
\(= 0.67 \text{M}\)