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

We are given the molarity of NaHSO4 solution, which is 0.10 M. To convert it to molality, we need the mass of the solvent. We can calculate it using the density formula and considering a 1-liter solution. Since the density of the solution is given as 1.002 g/mL or 1002 g/L, we can find the mass of the solute and solvent as follows: Mass of 1 liter solution = 1002 g Since Molarity = moles of solute / volume of solution in liters, 0.10 M NaHSO4 solution contains 0.10 moles of NaHSO4 in 1 liter of solution. Mass of solute (NaHSO4) = moles * molar mass = 0.10 moles * 120 g/mol (rounded) Mass of solute (NaHSO4) = 12 g Now we can calculate the mass of the solvent (water): Mass of solvent (water) = Mass of solution - Mass of solute Mass of solvent (water) = 1002 g - 12 g = 990 g Molality = moles of solute / mass of solvent in kg Molality = 0.10 moles / (990 g * 1 kg/1000 g) = 0.101 mol/kg

The molal boiling point elevation constant (Kb) for water is generally given and has a value of 0.512 °C/mol/kg. We can use this value in the boiling point elevation formula.

The boiling point elevation formula is given by: ΔTb = Kb × molality × i where ΔTb = boiling point elevation Kb = molal boiling point elevation constant of solvent molality = molality of the solution i = Van't Hoff factor (number of particles produced by solute in solution) We know that NaHSO4 will dissociate into Na+ and HSO4- ions. So, the Van't Hoff factor (i) will be 2. Now we can calculate the boiling point elevation: ΔTb = 0.512 °C/mol/kg × 0.101 mol/kg × 2 = 0.1035 °C

Since we have calculated the boiling point elevation, we can now find the boiling point of the NaHSO4 solution. The normal boiling point of pure water is 100 °C. Adding the boiling point elevation, we get the boiling point of the solution: Boiling point of solution = normal boiling point of water + ΔTb Boiling point of solution = 100 °C + 0.1035 °C ≈ 100.1 °C Therefore, the boiling point of the 0.10 M NaHSO4 solution is approximately 100.1 °C.