1.24 g of \(AX_2\) (molar mass 124 g mol\(^{-1}\)) is dissolved in 1 kg of water to form a solution with boiling point of 100.105°C, while 2.54 g of AY_2 (molar mass 250 g mol\(^{-1}\)) in 2 kg of water constitutes a solution with a boiling point of 100.026°C. \(Kb(H)_2\)\(\text(O)\) = 0.52 K kg mol\(^{-1}\). Which of the following is correct?
To solve this problem, we will use the concept of boiling point elevation. The formula for boiling point elevation is given by:
\[\Delta T_b = i \cdot K_b \cdot m\]
where \(\Delta T_b\) is the elevation in boiling point, \(i\) is the van 't Hoff factor (ionization number), \(K_b\) is the ebullioscopic constant, and \(m\) is the molality of the solution.
For AX₂:
Dissolved mass \(= 1.24 \, \text{g}\)
Molar mass \(= 124 \, \text{g/mol}\)
Molality (\(m\)) is calculated as:
\[m = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}} \times \frac{1}{\text{mass of solvent (kg)}} = \frac{1.24 \, \text{g}}{124 \, \text{g/mol}} \times \frac{1}{1 \, \text{kg}} = 0.01 \, \text{mol/kg}\]
For AY₂:
Dissolved mass \(= 2.54 \, \text{g}\)
Molar mass \(= 250 \, \text{g/mol}\)
Molality (\(m\)) is:
\[m = \frac{2.54 \, \text{g}}{250 \, \text{g/mol}} \times \frac{1}{2 \, \text{kg}} = 0.00508 \, \text{mol/kg}\]
Given \(K_b = 0.52 \, \text{K kg/mol}\).
For AX₂:
Boiling point change \(\Delta T_b = 100.105 - 100 = 0.105 \, °\text{C}\)
\(i \cdot 0.52 \cdot 0.01 = 0.105\)
\[i = \frac{0.105}{0.52 \times 0.01} = 20.192\]
The value \(i = 1\) indicates no ionization, meaning AX₂ is unionized.
For AY₂:
Boiling point change \(\Delta T_b = 100.026 - 100 = 0.026 \, °\text{C}\)
\(i \cdot 0.52 \cdot 0.00508 = 0.026\)
\[i = \frac{0.026}{0.52 \times 0.00508} \approx 0.985\]
This \(i \approx 2\) suggests AY₂ is ionized, breaking into AY and 2Y⁻ (fully ionized).
Given calculations indicate that AX₂ is completely unionized while AY₂ is fully ionized, which corresponds to the statement: "AX₂ is completely unionised while AY₂ is fully ionised."
The cause for deviation from Raoult’s law in the colligative properties of non-ideal solutions lies in the nature of interactions at the molecular level. These properties show deviations from Raoult’s law due to difference in interactions between solute–solvent, solute–solute and solvent–solvent. Some liquids on mixing form azeotropes which are binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation. There are two types of azeotropes called minimum boiling azeotrope and maximum boiling azeotrope. (a) Pure ethanol cannot be prepared by fractional distillation of ethanol–water mixture. Comment.
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is: