Question

5.17 m nahco₃ solution (density of solution = 1.19 \\(\frac{g}{ml}\\)). be sure your answer has the correct number of significant digits.

Explanation:

Step1: Define Molarity (M) and Molality (m)

Molarity (M) is moles of solute per liter of solution. Molality (m) is moles of solute per kilogram of solvent. Let's assume we have 1 L (1000 mL) of the solution.

Step2: Calculate mass of solution

Density = $\frac{\text{mass}}{\text{volume}}$, so mass of solution = density × volume.
Density = $1.19 \frac{\text{g}}{\text{mL}}$, volume = 1000 mL.
Mass of solution = $1.19 \frac{\text{g}}{\text{mL}} × 1000 \text{mL} = 1190 \text{g}$.

Step3: Calculate moles of solute ($\text{NaHCO}_3$)

Molarity (M) = 5.17 M, so moles of solute = M × volume (in L) = $5.17 \text{mol/L} × 1 \text{L} = 5.17 \text{mol}$.

Step4: Calculate mass of solute

Molar mass of $\text{NaHCO}_3$: Na (22.99) + H (1.01) + C (12.01) + 3O (16.00×3) = $22.99 + 1.01 + 12.01 + 48.00 = 84.01 \text{g/mol}$.
Mass of solute = moles × molar mass = $5.17 \text{mol} × 84.01 \text{g/mol} ≈ 434.33 \text{g}$.

Step5: Calculate mass of solvent

Mass of solvent = mass of solution - mass of solute = $1190 \text{g} - 434.33 \text{g} = 755.67 \text{g} = 0.75567 \text{kg}$.

Step6: Calculate molality (m)

Molality (m) = $\frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{5.17 \text{mol}}{0.75567 \text{kg}} ≈ 6.84 \text{m}$.

Answer:

The molality of the $\text{NaHCO}_3$ solution is approximately $\boxed{6.84 \text{m}}$ (or 6.84 mol/kg).