Question

The boiling point of water in a $01$ molal silver nitrate solution (solution $A$ ) is $x ^{\circ} C$ To this solution $A$, an equal volume of $01$ molal aqueous barium chloride solution is added to make a new solution $B$ The difference in the boiling points of water in the two solutions $A$ and $B$ is $y \times 10^{-2}{ }^{\circ} C$ (Assume: Densities of the solutions $A$ and $B$ are the same as that of water and the soluble salts dissociate completely Use: Molal elevation constant (Ebullioscopic Constant), $K _{ b }=05 \,K \,kg\, mol ^{-1}$; Boiling point of pure water as $100^{\circ} C$)
The value of $x$ is ______$^{\circ}C$

Answer 1

Correct Answer: 100.1

Step 1: Understanding the problem setup
We are given the following information:
- A solution \( A \) is a \( 0.1 \) molal silver nitrate solution.
- The boiling point of solution \( A \) is denoted as \( x \) degrees Celsius.
- An equal volume of \( 0.1 \) molal aqueous barium chloride solution is added to solution \( A \) to form solution \( B \).
- The difference in the boiling points of the two solutions \( A \) and \( B \) is denoted as \( y \times 10^{-2} \) degrees Celsius.
- The molal elevation constant (ebullioscopic constant) \( K_b = 0.5 \, \text{K} \, \text{kg/mol} \).
- The boiling point of pure water is \( 100^{\circ} \text{C} \).
Step 2: Using the formula for boiling point elevation
The boiling point elevation \( \Delta T_b \) is given by the formula:
\[ \Delta T_b = K_b \cdot m \cdot i \] where: - \( \Delta T_b \) is the change in the boiling point of the solution, - \( K_b \) is the ebullioscopic constant (given as 0.5 \( \text{K} \, \text{kg/mol} \)), - \( m \) is the molality of the solution, - \( i \) is the van't Hoff factor, which represents the number of particles into which the solute dissociates.
For silver nitrate \( \text{AgNO}_3 \), the van't Hoff factor \( i = 2 \) because it dissociates into two ions: \( \text{Ag}^+ \) and \( \text{NO}_3^- \).
For barium chloride \( \text{BaCl}_2 \), the van't Hoff factor \( i = 3 \) because it dissociates into three ions: \( \text{Ba}^{2+} \) and two \( \text{Cl}^- \).
Step 3: Calculating the boiling point of solution \( A \)
For solution \( A \) (silver nitrate), the molality \( m = 0.1 \, \text{mol/kg} \), and the van't Hoff factor \( i = 2 \). Using the formula for boiling point elevation:
\[ \Delta T_b = K_b \cdot m \cdot i = 0.5 \cdot 0.1 \cdot 2 = 0.1 \, \text{K} \] Therefore, the boiling point of solution \( A \) is:
\[ \text{Boiling point of solution } A = 100^{\circ} \text{C} + 0.1^{\circ} \text{C} = 100.1^{\circ} \text{C} \] Hence, the value of \( x \) is \( 100.1^{\circ} \text{C} \).
Step 4: Conclusion
Therefore, the value of  x is 100.1.