Question

What is the boiling point of a \(0.10 \mathrm{M}\) solution of \(\mathrm{NaHSO}_{4}\) if the solution has a density of \(1.002 \mathrm{~g} / \mathrm{mL} ?\)

Short answer

The boiling point of a 0.10 M NaHSO4 solution with a density of 1.002 g/mL is 100.052 °C.

Step-by-step solution

Write down the boiling point elevation formula

The boiling point elevation formula is given by: \[ΔT_b = K_b \cdot m\] Where \(ΔT_b\) is the boiling point elevation, \(K_b\) is the ebullioscopic constant for water (0.512 °C/m), and \(m\) is the molality of the solution.

Convert molarity to molality

To convert molarity to molality, first, we need to determine the mass of the solvent (water). Using the density and the volume of the solution, we can find the mass of the water: 1.002 g/mL = mass of solution / 1 mL of solution Since we have a 0.10 M solution, there are 0.10 moles of NaHSO4 in 1 L (1000 mL) of solution. The mass of NaHSO4 in 1 L of solution can be calculated using the molecular weight of NaHSO4, which is 120.06 g/mol: mass of NaHSO4 = (0.10 moles/L) × (120.06 g/mol) = 12.006 g Now, we can find the mass of water in 1 L of solution: mass of water = mass of solution - mass of NaHSO4 = (1.002 g/mL × 1000 mL) - 12.006 g = 990.00 g Since molality is defined as moles of solute per kilogram of solvent, we can find the molality of the solution: molality = (0.10 moles / 0.990 kg) = 0.101 mol/kg

Calculate the boiling point elevation

Using the boiling point elevation formula, we now can calculate the boiling point elevation: \(ΔT_b = K_b \cdot m\) \(ΔT_b =\) 0.512 °C/m × 0.101 mol/kg = 0.0517 °C

Find the new boiling point of the solution

Finally, we add the boiling point elevation to the normal boiling point of water (100 °C) to find the boiling point of the 0.10 M NaHSO4 solution: boiling point of the solution = 100 °C + 0.0517 °C = 100.052 °C So the boiling point of a 0.10 M NaHSO4 solution with a density of 1.002 g/mL is 100.052 °C.