Question

3.0 molal NaOH solution has a density of 1.11 g/ml. The molarity of the solution is

• A. 2.97
• B. 3.05
• C. 3.64
• D. 3.050

Answer 1

Answer: (A)

2.97

Answer 2

Let's solve the problem using the definitions of molality and molarity.

Molality (m) is defined as the number of moles of solute per kilogram of solvent. Given molality = 3.0 molal. This means there are 3.0 moles of NaOH solute in 1 kg (1000 g) of solvent.

The solute is NaOH. The molar mass of NaOH ($M_o$) is the sum of the atomic masses of Na, O, and H. $M_o(\text{NaOH}) = \text{Atomic mass of Na} + \text{Atomic mass of O} + \text{Atomic mass of H}$ $M_o(\text{NaOH}) = 23.0 \text{ g/mol} + 16.0 \text{ g/mol} + 1.0 \text{ g/mol} = 40.0 \text{ g/mol}$.

Mass of solute = number of moles of solute $\times$ molar mass of solute Mass of NaOH = 3.0 mol $\times$ 40.0 g/mol = 120 g.

Mass of solvent = 1000 g (from the definition of 3.0 molal solution).

Mass of solution = Mass of solute + Mass of solvent Mass of solution = 120 g + 1000 g = 1120 g.

Density of the solution is given as 1.11 g/ml. Density (d) = $\frac{\text{Mass of solution}}{\text{Volume of solution}}$. Volume of solution = $\frac{\text{Mass of solution}}{\text{Density}}$ Volume of solution = $\frac{1120 \text{ g}}{1.11 \text{ g/ml}} = \frac{1120}{1.11} \text{ ml}$.

Molarity (M) is defined as the number of moles of solute per liter of solution. Number of moles of solute = 3.0 mol. Volume of solution in liters = $\frac{\text{Volume in ml}}{1000}$ Volume of solution in liters = $\frac{1120/1.11}{1000} \text{ L} = \frac{1120}{1.11 \times 1000} \text{ L} = \frac{1120}{1110} \text{ L}$.

Molarity (M) = $\frac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}$ M = $\frac{3.0 \text{ mol}}{\frac{1120}{1110} \text{ L}} = \frac{3.0 \times 1110}{1120} \text{ mol/L}$. M = $\frac{3 \times 111}{112} \text{ mol/L} = \frac{333}{112} \text{ mol/L}$.

Now, we calculate the numerical value: $\frac{333}{112} \approx 2.973214...$

Comparing this value with the given options, the calculated value 2.9732 is closest to 2.97.

Alternatively, we can use the formula relating molarity (M), molality (m), density (d), and molar mass of solute ($M_o$): $M = \frac{1000 \times m \times d}{1000 + m \times M_o}$ Given: m = 3.0 mol/kg, d = 1.11 g/ml, $M_o$ = 40.0 g/mol. $M = \frac{1000 \times 3.0 \times 1.11}{1000 + 3.0 \times 40.0}$ $M = \frac{3330}{1000 + 120.0}$ $M = \frac{3330}{1120.0}$ $M = \frac{3330}{1120} = \frac{333}{112} \approx 2.9732...$

Rounding to two decimal places, the molarity is 2.97 M.