Arrange the following in increasing order of their molarity when their equal weights are dissolved in 250 mL of water.

(A) glucose ($180\, g\, mol^{-1}$)
(B) sodium carbonate ($106\, g\, mol^{-1}$)
(C) sodium hydroxide ($40\, g\, mol^{-1}$)
(D) acetic acid ($60\, g\, mol^{-1}$)

Choose the correct answer from the options given below:

(A), (B), (D), (C)

The correct answer is Option (3) → (A), (B), (D), (C)

Molarity $M = \frac{\text{moles of solute}}{\text{volume of solution in L}}$

Given equal weights, the compound with lower molar mass will have more moles and hence higher molarity.

• Molar masses:
• Glucose (A) = 180 g/mol → moles = 1/180
• Sodium carbonate (B) = 106 g/mol → moles = 1/106
• Sodium hydroxide (C) = 40 g/mol → moles = 1/40
• Acetic acid (D) = 60 g/mol → moles = 1/60
• Increasing order of moles (and molarity):

$\text{Glucose (A)} < \text{Na₂CO₃ (B)} < \text{Acetic acid (D)} < \text{NaOH (C)}$