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 rank the molarity of solutions containing 25 grams of each compound in the same volume of water, we must determine the number of moles for each. Molarity (\( M \)) is the ratio of moles of solute to liters of solution. Given a constant volume, molarity increases with the number of moles.

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:
   \[ \text{Moles of NaOH} = \frac{25 \, \text{g}}{40.0 \, \text{g/mol}} = 0.625 \, \text{mol} \] \[ \text{Moles of LiOH} = \frac{25 \, \text{g}}{23.94 \, \text{g/mol}} \approx 1.043 \, \text{mol} \] \[ \text{Moles of KOH} = \frac{25 \, \text{g}}{56.10 \, \text{g/mol}} \approx 0.446 \, \text{mol} \] \[ \text{Moles of Al(OH)}_3 = \frac{25 \, \text{g}}{77.98 \, \text{g/mol}} \approx 0.321 \, \text{mol} \] \[ \text{Moles of B(OH)}_3 = \frac{25 \, \text{g}}{61.81 \, \text{g/mol}} \approx 0.404 \, \text{mol} \]
3. Relate Moles to Molarity:
   With equal volumes, molarity is directly proportional to moles. Higher mole values correspond to higher molarities.
   \[ \text{LiOH} \approx 1.043 \, \text{mol} \quad \text{NaOH} = 0.625 \, \text{mol} \quad \text{KOH} \approx 0.446 \, \text{mol} \] \[ \text{B(OH)}_3 \approx 0.404 \, \text{mol} \quad \text{Al(OH)}_3 \approx 0.321 \, \text{mol} \]
4. Increasing Order of Molarity:
   Based on the calculated moles:
   \[ \text{Al(OH)}_3 (D) < \text{B(OH)}_3 (E) < \text{KOH} (C) < \text{NaOH} (A) < \text{LiOH} (B) \] Hence, the increasing molarity order is: \[ (D) < (E) < (C) < (A) < (B) \]

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