Question

1 litre decinormal solution of KCl is prepared. Half of the solution is converted into centinormal and added it into left decinormal solution. Then

• A. Number of milli equivalents of KCl do not change
• B. Number of milli moles of KCl are reduced by 1/5
• C. Normality of the final solution becomes 0.001 N
• D. Molarity of the final solution becomes nearly 0.018 M

Answer 1

Answer: (A), (D)

A, D

Answer 2

The problem describes a process involving the dilution and mixing of a KCl solution. We need to determine the correct statement(s) among the given options.

1. Initial State:

• Volume of KCl solution ($V_{initial}$) = 1 litre = 1000 mL
• Normality of KCl solution ($N_{initial}$) = decinormal = 0.1 N
• Number of milliequivalents of KCl initially ($meq_{initial}$) = $N_{initial} \times V_{initial} = 0.1 \text{ N} \times 1000 \text{ mL} = 100 \text{ meq}$.
• Since KCl is a 1:1 electrolyte ($K^+ + Cl^-$), its n-factor is 1. Therefore, Normality = Molarity, and milliequivalents = millimoles.
• Number of millimoles of KCl initially ($mmol_{initial}$) = 100 mmol.

2. Splitting and Conversion: The initial solution is divided into two equal halves:

• Part A: 500 mL of 0.1 N KCl solution.
  • Number of milliequivalents in Part A ($meq_A$) = $0.1 \text{ N} \times 500 \text{ mL} = 50 \text{ meq}$.
• Part B (the "left decinormal solution"): 500 mL of 0.1 N KCl solution.
  • Number of milliequivalents in Part B ($meq_B$) = $0.1 \text{ N} \times 500 \text{ mL} = 50 \text{ meq}$.

Part A is "converted into centinormal". This means Part A is diluted so its normality becomes 0.01 N (centinormal). Let the new volume of Part A be $V_A'$. Using the dilution formula $N_1V_1 = N_2V_2$: $0.1 \text{ N} \times 500 \text{ mL} = 0.01 \text{ N} \times V_A'$ $V_A' = \frac{0.1 \times 500}{0.01} = 5000 \text{ mL}$. The number of milliequivalents in Part A remains 50 meq after dilution, as dilution only changes concentration and volume, not the amount of solute.

3. Mixing: The diluted Part A ($V_A' = 5000 \text{ mL}$, $N_A' = 0.01 \text{ N}$) is added to Part B ($V_B = 500 \text{ mL}$, $N_B = 0.1 \text{ N}$).

• Total volume of the final solution ($V_{final}$) = $V_A' + V_B = 5000 \text{ mL} + 500 \text{ mL} = 5500 \text{ mL}$.
• Total number of milliequivalents in the final solution ($meq_{final}$) = $meq_A + meq_B = 50 \text{ meq} + 50 \text{ meq} = 100 \text{ meq}$.

4. Evaluating the Options:

(A) Number of milli equivalents of KCl do not change Initial milli equivalents = 100 meq. Final milli equivalents = 100 meq. Since no KCl was added or removed from the system, the total number of milliequivalents remains constant. This statement is TRUE.

(B) Number of milli moles of KCl are reduced by 1/5 For KCl, millimoles = milliequivalents (n-factor = 1). Initial millimoles = 100 mmol. Final millimoles = 100 mmol. The number of millimoles does not change. This statement is FALSE.

(C) Normality of the final solution becomes 0.001 N Normality of final solution ($N_{final}$) = $\frac{meq_{final}}{V_{final} \text{ (in mL)}} = \frac{100 \text{ meq}}{5500 \text{ mL}} = \frac{1}{55} \text{ N} \approx 0.01818 \text{ N}$. This statement is FALSE.

(D) Molarity of the final solution becomes nearly 0.018 M Since Normality = Molarity for KCl, the molarity of the final solution ($M_{final}$) = $N_{final} = \frac{1}{55} \text{ M} \approx 0.01818 \text{ M}$. This value is nearly 0.018 M. This statement is TRUE.