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$= \frac{0.056\text{mol}}{83.33\times10^{-3} \text{L}}$
$= 0.67 \text{M}$