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 $gm{{L}^{-1}}$, then what shall be the molarity of the solution?

Answer 1

Hint: We are asked to find the mole fraction, molarity and molality of a glucose solution. The terms molarity and molality are a way to express the concentration of a solution and it can be obtained dividing the moles of solute by the kilograms or litres of solvent.

Complete step by step solution:
-It’s given in the question that the glucose present in the water is 10 % ${w}/{w}\;$. The 10 % ${w}/{w}\;$ solution of glucose in water means that there is 10 grams of glucose present in 100 grams of the solution which we take. That is in a total 100 g of solution1 10 g will be glucose and the remaining (100− 10) i.e. 90g will be water.
- The molecular formula of glucose is ${{C}_{6}}{{H}_{12}}{{O}_{6}}$ .From this molecular formula the molar mass of glucose can be found as follows
\[Molar\text{ }mass\text{ }of\text{ }glucose\left( {{C}_{6}}{{H}_{12}}{{O}_{6}} \right)=\left( 6\times 12 \right)+\left( 12\times 1 \right)+\left( 6\times 16 \right)=180gmo{{l}^{-1}}\] Thus the molar mass of glucose is 180 gram per mole. As we know. Number of moles in a sample can be obtained by dividing the given mass (10 g) by the molar mass (180$gmo{{l}^{-1}}$).Hence the number of moles of glucose can be found as follows
\[Number\text{ }of\text{ }moles\text{ }of\text{ }glucose=\dfrac{10g}{180gmo{{l}^{-1}}}=0.056mol\]
Molality is a way to express the concentration of a solution and it can be obtained dividing the moles of solute (0.056 mol) by the kilograms of solvent (90g=0.09kg).
\[Molality\text{ }of\text{ }solution=\dfrac{0.056mol}{0.09kg}=0.62m\]
The number of moles of water in the solution (given mass=90 g and molar mass of water=18 $gmo{{l}^{-1}}$ ) can be given as follows
 \[Number\text{ }of\text{ }moles\text{ }of\text{ }water=\dfrac{90g}{18gmo{{l}^{-1}}}=5mol\]
The term mole fraction can be defined as the unit of the amount of a constituent usually expressed in moles, divided by the total amount of all constituents in a mixture and the mole fraction of glucose can thus be written as follows
\[Mole\text{ }fraction\text{ }of\text{ }glucose=\dfrac{0.056}{\left( 0.056+5 \right)}=0.011\]
As we know the sum of mole fractions of solute and solvent is 1 and thus the mole fraction of water can be found from the mole fraction of glucose as follows
\[Mole\text{ }fraction\text{ }of\text{ }water=1-0.011=0.989\]
The density of the solution is given as 1.2$gm{{L}^{-1}}$ and from this the volume of the 100 g solution can be written as follows
\[volume\text{ }of\text{ }the\text{ }100\text{ }g\text{ }solution=\dfrac{100g}{1.2m{{L}^{-1}}}=83.33mL=83.33\times {{10}^{-3}}L\]
As in the case of molality, molarity is also a way to express the concentration of a solution and is defined as the moles of a solute (0.056 mol) per liters of a solution ($83.33\times {{10}^{-3}}L$). Therefore the molarity of the solution can be written as
\[Molarity\text{ }of\text{ }the\text{ }solution=\dfrac{0.0.056mol}{83.33\times {{10}^{-3}}L}=0.67M\]

Therefore the molarity of the solution is 0.67 M.

Note: Students might confuse between molarity and molality. Molarity is the number of moles of solute per litre and molality is the number of moles per 1 kg of solvent. Molarity is temperature and volume dependent whereas molality depends on mass but independent of temperature.