Question

The molarity of 10% (w/w) aqueous NaOH solution (density 1.11 g mL\(^{-1}\)) is:

• A. 2.50 M
• B. 3.25 M
• C. 2.78 M
• D. 1.52 M

Hint:

For converting percentage composition into molarity, first calculate the mass of solution using density, then determine the mass of solute and finally apply: \[ M=\frac{\text{moles of solute}}{\text{volume of solution in litres}} \]

Answer 1

The Correct Option is C

Step 1: Molarity formula.
M = mass of solute (g) / [molar mass (g/mol) × volume of solution (L)].

Step 2: Finding solute mass in 1 L.
Density = 1.11 g/mL, so mass of 1 L solution = 1000 mL × 1.11 g/mL = 1110 g. 10% w/w NaOH means NaOH mass = 0.10 × 1110 g = 111 g.

Step 3: Moles of NaOH.
Molar mass of NaOH = 40 g/mol, thus moles = 111 g / 40 g/mol = 2.775 mol.

Step 4: Determining molarity.
Volume = 1 L, so molarity = 2.775 mol / 1 L = 2.775 M.

Step 5: Rounding appropriately.
Molarity ≈ 2.78 M.

Step 6: Conclusion.
Therefore, the solution's molarity is 2.78 M.