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: Write the formula for molarity.
\[ M = \frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)} \times \text{volume of solution (L)}} \]

Step 2: Determine mass of solute in 1 L solution.
Density = 1.11 g/mL → mass of 1 L solution = \(1000 \text{ mL} \times 1.11 \text{ g/mL} = 1110 \text{ g}\)
10% w/w NaOH → mass of NaOH = 0.10 × 1110 g = 111 g

Step 3: Calculate number of moles of NaOH.
Molar mass NaOH = 40 g/mol → moles = 111 g 40 g/mol = 2.775 mol

Step 4: Determine solution volume.
Volume = 1 L (as given) → molarity = 2.775 mol 1 L = 2.775 M

Step 5: Round to appropriate significant figures.
\(\text{Molarity} \approx 2.78\) M

Step 6: Conclusion.
Hence, the molarity of the solution is 2.78 M.