Question

18% wt/vol glucose aq solution is given if density of sol is 1.1g/ml then find
1.molarity
2.milality
3.mole fraction of both components​

Answer 1

18% Wt/Volume glucose aqueous solution means that 18 gm of Glucose is present in 100 cm³ of solution.

We have been given the density of the solution to be 1.1gm/ml.

Density = $\dfrac{Mass}{Volume}$

Substitute the values:-

1.1 = $\dfrac{Mass}{100}$

Mass = 100 x 1.1

Mass = 110 gm

Mass of solution is 110 gm.

Mass of water in solution is 110 - 18 = 92 gm

1. Molarity = $\dfrac{Moles of Solute}{Volume of Solution}\times 1000$

= $\dfrac{18}{100\times180} \times 1000$

= 1 M

2. Molality = $\dfrac{Moles of Solute}{Mass of Solvent}\times 1000$

= $\dfrac{18}{92\times180} \times 1000$

= 1.08 m

3. Mole Fraction:-

a) Glucose = $\dfrac{Moles of Glucose}{Moles of Glucose + Moles of Water}$

= $\dfrac{\dfrac{18}{180}}{\dfrac{18}{180} + \dfrac{92}{18}}$

= $\dfrac{\dfrac{1}{10}}{\dfrac{1}{10}+\dfrac{46}{9}}$

= $\dfrac{\dfrac{1}{10}}{\dfrac{9+460}{90}}$

= $\dfrac{\dfrac{1}{10}}{\dfrac{469}{90}}$

= $\dfrac{1}{10} \times \dfrac {90}{469}$

= $\dfrac{9}{469}$

b) Water = 1 - $\dfrac{9}{469}$

= $\dfrac{460}{469}$