Question

The increasing order of molarity of 25 gm each of:
(A) \( \text{NaOH} \)
(B) \( \text{LiOH} \) 
(C) \( \text{KOH} \) 
(D) \( \text{Al(OH)}_3 \) 
(E) \( \text{B(OH)}_3 \)
in the same volume of water is:
Choose the correct answer from the options given below:

• (D) < (E) < (C) < (A) < (B)
• (D) < (C) < (B) < (A) < (E)
• (B) < (A) < (C) < (E) < (D)
• (B) < (C) < (D) < (A) < (E)

Hint:

Solubility and dissociation determine the molarity of a solution.

Answer 1

The Correct Option is A

To determine the increasing order of molarity for 25 grams of each compound dissolved in the same volume of water, we need to calculate the number of moles of each substance. Molarity (\( M \)) is defined as the number of moles of solute per liter of solution. Since the volume of water is the same for all, the substance with the highest number of moles will have the highest molarity.

1. Calculate Molar Masses:
   \[ \text{NaOH: } \text{Na}(23.0) + \text{O}(16.0) + \text{H}(1.0) = 40.0 \, \text{g/mol} \] \[ \text{LiOH: } \text{Li}(6.94) + \text{O}(16.0) + \text{H}(1.0) = 23.94 \, \text{g/mol} \] \[ \text{KOH: } \text{K}(39.10) + \text{O}(16.0) + \text{H}(1.0) = 56.10 \, \text{g/mol} \] \[ \text{Al(OH)}_3: \text{Al}(26.98) + 3 \times [\text{O}(16.0) + \text{H}(1.0)] = 26.98 + 51.0 = 77.98 \, \text{g/mol} \] \[ \text{B(OH)}_3: \text{B}(10.81) + 3 \times [\text{O}(16.0) + \text{H}(1.0)] = 10.81 + 51.0 = 61.81 \, \text{g/mol} \]
2. Calculate Number of Moles for Each Compound:
   \[ \text{Moles of NaOH} = \frac{25 \, \text{g}}{40.0 \, \text{g/mol}} = 0.625 \, \text{mol} \] \[ \text{Moles of LiOH} = \frac{25 \, \text{g}}{23.94 \, \text{g/mol}} \approx 1.043 \, \text{mol} \] \[ \text{Moles of KOH} = \frac{25 \, \text{g}}{56.10 \, \text{g/mol}} \approx 0.446 \, \text{mol} \] \[ \text{Moles of Al(OH)}_3 = \frac{25 \, \text{g}}{77.98 \, \text{g/mol}} \approx 0.321 \, \text{mol} \] \[ \text{Moles of B(OH)}_3 = \frac{25 \, \text{g}}{61.81 \, \text{g/mol}} \approx 0.404 \, \text{mol} \]
3. Determine Molarity:
   Assuming the same volume of water for each solution, the number of moles directly corresponds to molarity. Therefore, higher moles imply higher molarity.
   \[ \text{LiOH} \approx 1.043 \, \text{mol} \quad \text{NaOH} = 0.625 \, \text{mol} \quad \text{KOH} \approx 0.446 \, \text{mol} \] \[ \text{B(OH)}_3 \approx 0.404 \, \text{mol} \quad \text{Al(OH)}_3 \approx 0.321 \, \text{mol} \]
4. Arrange in Increasing Order of Molarity:
   From the calculations:
   \[ \text{Al(OH)}_3 (D) < \text{B(OH)}_3 (E) < \text{KOH} (C) < \text{NaOH} (A) < \text{LiOH} (B) \] Therefore, the increasing order of molarity is: \[ (D) < (E) < (C) < (A) < (B) \]

Therefore, the increasing order of molarity is (D) < (E) < (C) < (A) < (B).