1.02 g of the following substances are dissolved in 100 mL of water. Arrange them in increasing order of their molarity:
(A) \( NaCl \)
(B) \( KCl \)
(C) \( MgSO_4 \)
(D) \( NaOH \)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C), (B), (A), (D)
Molarity is the number of moles of solute per litre of solution:
\( \displaystyle M = \frac{\text{moles of solute}}{\text{volume in L}} = \frac{w/M_{\text{solute}}}{V} \)
Here every solution has the same mass (1.02 g) in the same volume (100 mL = 0.1 L), so the molarity depends only on the molar mass: the larger the molar mass, the fewer the moles and the lower the molarity.
(A) \( \ce{NaCl} \): 23 + 35.5 = 58.5 g mol−1; \( \frac{1.02}{58.5} = 0.0174 \) mol, so M = 0.174 M.
(B) \( \ce{KCl} \): 39 + 35.5 = 74.5 g mol−1; \( \frac{1.02}{74.5} = 0.0137 \) mol, so M = 0.137 M.
(C) \( \ce{MgSO4} \): 24 + 32 + 64 = 120 g mol−1; \( \frac{1.02}{120} = 0.0085 \) mol, so M = 0.085 M.
(D) \( \ce{NaOH} \): 23 + 16 + 1 = 40 g mol−1; \( \frac{1.02}{40} = 0.0255 \) mol, so M = 0.255 M.
In increasing order of molarity:
\( \displaystyle \ce{MgSO4}\,(0.085) \lt \ce{KCl}\,(0.137) \lt \ce{NaCl}\,(0.174) \lt \ce{NaOH}\,(0.255) \)
Shortcut: No division is needed. Increasing molarity is the order of decreasing molar mass: 120 (C) > 74.5 (B) > 58.5 (A) > 40 (D).