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.

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

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

Given:

Mass of each solute = 2.5 g

Volume of solution = 500 mL = 0.5 L

Molarity (M) = (Number of moles of solute) / (Volume in litres)

Number of moles = mass / molar mass

Calculations:

(A) C₆H₁₂O₆ (Glucose), Molar mass = 180 g/mol

Moles = 2.5 / 180 = 0.01389 mol

Molarity = 0.01389 / 0.5 = 0.02778 M

(B) NH₂CONH₂ (Urea), Molar mass = 60 g/mol

Moles = 2.5 / 60 = 0.04167 mol

Molarity = 0.04167 / 0.5 = 0.08333 M

(C) NaCl, Molar mass = 58.5 g/mol

Moles = 2.5 / 58.5 ≈ 0.04274 mol

Molarity = 0.04274 / 0.5 ≈ 0.08548 M

(D) MgSO₄, Molar mass = 120 g/mol

Moles = 2.5 / 120 = 0.02083 mol

Molarity = 0.02083 / 0.5 = 0.04167 M

Molarity values:

• (A) Glucose → 0.02778 M
• (D) MgSO₄ → 0.04167 M
• (B) Urea → 0.08333 M
• (C) NaCl → 0.08548 M

Increasing order of molarity: (A) < (D) < (B) < (C)