2.5 g of the following substances are dissolved in 500 ml of water. Arrange the resultant solutions in increasing order of their molarity.
(A) \( C_6H_{12}O_6 (180\, g\, mol^{-1}) \)
(B) \( NH_2CONH_2 (60\, g\, mol^{-1}) \)
(C) \( NaCl (58.5\, g\, mol^{-1}) \)
(D) \( MgSO_4 (120\, g\, mol^{-1}) \)
Choose the correct order from the options given below.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (D), (B), (C)
Molarity is the number of moles of solute per litre of solution:
\( \displaystyle M = \frac{\text{mass}/M_{\text{solute}}}{V\ (\mathrm{L})} \)
Here every solution has 2.5 g of solute in 0.5 L, so
\( \displaystyle M = \frac{2.5/M_{\text{solute}}}{0.5} = \frac{5}{M_{\text{solute}}} \)
The same mass in the same volume gives a larger molarity for a smaller molar mass:
\( \displaystyle \text{(A) Glucose: } \frac{5}{180} = 0.028\ \mathrm{M} \)
\( \displaystyle \text{(B) Urea: } \frac{5}{60} = 0.083\ \mathrm{M} \)
\( \displaystyle \text{(C) NaCl: } \frac{5}{58.5} = 0.085\ \mathrm{M} \)
\( \displaystyle \text{(D) } \ce{MgSO4}\text{: } \frac{5}{120} = 0.042\ \mathrm{M} \)
Molarity counts formula units of the solute, so the dissociation of NaCl and \( \ce{MgSO4} \) into ions does not change it; dissociation matters only for colligative properties.
Shortcut: No division is needed. Arrange the molar masses in decreasing order, 180 > 120 > 60 > 58.5, and that is the increasing order of molarity: (A), (D), (B), (C).