Question

How many g of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ are there in 100 ml of 0.15 m solution of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$. The density of the solution is 1.5 g/ml. Also report the no. of $A{{l}^{3+}}$ ions in this weight.

Answer 1

Hint: Molality (m) is defined as the number of moles of solute per kilogram of the solvent. A commonly used unit for molality is mol/Kg. A solution of concentration 1 mol/Kg is also represented as 1molal. For example, 1.00 mol/Kg of $NaCl$ solution means that 1 mole (58.5 g) of $NaCl$ is dissolved in 1 Kg of water.

Complete step by step answer:
We have studied about the concepts of physical chemistry dealing with molarity, molality, normality etc. Let us see in detail about molality.
Given that,
Volume of solution (V) = 100 ml
Density of solution = 1.5 g/mol
Molality of solution = 0.15m
Mass of solution, w = density of the solution $\times $ volume of solution
\[\Rightarrow w = 1.5g/mol\times 100ml = 150g\]
Let,${{w}_{solute}}$ be the mass of solute, then $150-{{w}_{solute}}={{w}_{solvent}}$
Where,${{w}_{solvent}}$ be the mass of solvent.
Molar mass of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ is 342 g/mol
Therefore, molality of the solution can be found by using the formula,
\[m=\dfrac{{{w}_{solute}}}{\left( \dfrac{{{M}_{A{{l}_{2}}{{(S{{O}_{4}})}_{3}}}}}{{{w}_{solvent}}} \right)}\times 1000\]
- Now, by substituting the values, we have
\[0.15m=\dfrac{{{w}_{solute}}}{342(150-{{w}_{solvent}})}\times 1000\]
By solving this equation, we get weight of the solute as, ${{w}_{solute}}=7.32g$
This is nothing but the mass of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ is 7.32g
Thus, mass of $A{{l}^{3+}}$ ion = mass fraction of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$$\times $ mass of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$
- This implies that mass of $A{{l}^{3+}}$ ion will be equal to $\dfrac{2\times 27}{342}\times 7.32=1.156g$
Hence, 7.32g of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ are there in 100 ml of 0.15m solution of $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$. The density of the solution is 1.5g/ml, and the weight of $A{{l}^{3+}}$ is 1.156 g.

Note: Each method of expressing the concentration of the solutions has its own merits and demerits. Mass percent, ppm, mole fraction, and molality are independent of temperature whereas molarity is a function of temperature because molarity is expressed in volume which depends on temperature. But molality does not depend on temperature which is expressed in mass.