Question

A solution of glucose in water is labelled as 10% (\[\dfrac{w}{v}\]), what would be the molality and mole fraction of each component in the solution? If the density of the solution is $1.2\, gm{{L}^{-1}}$then what shall be the molarity of the solution?

Answer 1

Hint: For solving the given question you should have knowledge about the following terms Molality, Molarity and Mole fraction.
- Molarity (M) is defined as a number of moles of solute dissolved in one litre of solution.
- Molality (m) is defined as the number of moles of the solute per kilogram (kg) of the solvent.
- Mole fraction represents the number of moles of a particular component in a mixture divided by the total number of moles in the given mixture. It can be represented by X.

Complete step by step answer:
Mass by volume percentage (\[\dfrac{w}{v}\]) is the mass of solute dissolved in 100 mL of the solution.
Consider 100g of 10% solution of glucose whose conc. is 10% \[\dfrac{w}{w}\].
Mass of solution = 100g
Mass of glucose = 10g
Mass of solvent = 100 – 10 = 90g
Molar mass of glucose, ${{C}_{6}}{{H}_{12}}{{O}_{6}}=12\times 6+12\times 1+16\times 6$

$=72+12+96=180g/mol$

Molality of the solution is the number of moles of the solute per kilogram of the solvent.
 $Molality=\dfrac{Mass\,of\,Solute}{Molar\,Mass\,of\,Solute}\times\dfrac{1}{Mass\,of\,Solvent\,in\,Kg}$

= $\dfrac{10}{180\times 0.09kg}=0.617m$

Hence, the molality of the solution is 0.617m.

Moles of ${{H}_{2}}O,{{n}_{A}}=\dfrac{90}{18}=5$

Moles of glucose, ${{n}_{B}}=\dfrac{10}{180}=0.055$

Mole fraction of glucose, ${{x}_{B}}=\dfrac{{{n}_{B}}}{{{n}_{A}}+{{n}_{B}}}$

                                                  = $\dfrac{0.055}{5+0.055}=0.0108=0.01$

Hence, Mole fraction of glucose is 0.01
Mole fraction of water, ${{x}_{A}}=1-{{x}_{B}}=1-0.0108=0.989=0.99$
Hence, Mole fraction of water is 0.99
Density of solution is 1.2 g mol L as per given in question.

Therefore, volume of the solution $=\dfrac{Mass}{Density}=\dfrac{100}{1.2}=83.33mL$
Molarity(M) is a number of moles of solute dissolved in one litre of solution.

$Molarity=\dfrac{Mass\,of\,solute}{Molar\,mass\,of\,solute}\times \dfrac{1}{Volume\,of\,solution}$

= $\dfrac{10}{180}\times \dfrac{1000}{83.33}=0.67M$

Hence, Molarity of the solution is 0.67M.

Note: Don’t get confused in Molarity(M) and Molality(m). They are two different things. Molarity is defined as the number of moles of solute dissolved in one litre of solution whereas Molality is defined as the number of moles of the solute per kilogram (kg) of the solvent.