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:
Answer & explanation
Correct answer: option 3
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):
$\text{Glucose (A)} < \text{Na₂CO₃ (B)} < \text{Acetic acid (D)} < \text{NaOH (C)}$