Question

Calculate the number of moles of solute present in each of the following solutions: (a) \(185 \mathrm{~mL}\) of \(1.50 \mathrm{M}\) \(\mathrm{HNO}_{3}(a q)\), (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is \(1.25 \mathrm{~m} \mathrm{NaCl}\), (c) \(75.0 \mathrm{~g}\) of an aqueous solution that is \(1.50 \%\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) by mass.

Short answer

In the given solutions, (a) there are 0.2775 moles of HNO₃, (b) there are 0.06240 moles of NaCl, and (c) there are 0.00328 moles of sucrose (C₁₂H₂₂O₁₁).

Step-by-step solution

Solution (a) - Calculate moles using molarity and volume

Given the volume of the solution (V) is 185 mL and the molarity (M) is 1.50 M, we can calculate the number of moles of solute (n) using the formula n = MV, where n is the number of moles, M is the molarity, and V is the volume. First, convert the volume to liters: \(185 ~\mathrm{mL} \times \frac{1 ~\mathrm{L}}{1000 ~\mathrm{mL}} = 0.185 ~\mathrm{L}\) Now, we can apply the formula: \(n = MV = (1.50 ~\mathrm{M})(0.185 ~\mathrm{L}) = 0.2775 ~\mathrm{moles}\) So, there are 0.2775 moles of HNO₃ in the solution.

Solution (b) - Calculate moles using mass and molality

Given the mass of the solution (m_sol) is 50.0 mg and the molality (m) is 1.25 m, we need to convert mass into grams and then find the mass of the solute using molality, before finding the number of moles. First, convert mass to grams: \(50.0 ~\mathrm{mg} \times \frac{1 ~\mathrm{g}}{1000 ~\mathrm{mg}} = 0.0500 ~\mathrm{g}\) Next, we can set up the equation to determine the mass of the solute (m_solute) using molality and mass of the solvent (m_solvent): \(m = \frac{n}{m_\mathrm{solvent}}\), and \(m_\mathrm{solute} = n \times M_\mathrm{solute}\), where M_solute is the molar mass of NaCl (approximately 58.44 g/mol). We can combine these two equations: \(m = \frac{m_\mathrm{solute}}{M_\mathrm{solute} \times m_\mathrm{solvent}}\). Now solve for m_solute: \(m_\mathrm{solute} = m \times M_\mathrm{solute} \times m_\mathrm{solvent} = (1.25 ~\mathrm{m}) (58.44 ~\mathrm{g/mol}) (0.0500 ~\mathrm{g}) = 3.645 ~\mathrm{g}\) Finally, calculate the number of moles: \(n = \frac{m_\mathrm{solute}}{M_\mathrm{solute}} = \frac{3.645 ~\mathrm{g}}{58.44 ~\mathrm{g/mol}} = 0.06240 ~\mathrm{moles}\) So, there are 0.06240 moles of NaCl in the solution.

Solution (c) - Calculate moles using mass percentage and molar mass

Given the mass of the solution (m_sol) is 75.0 g and the mass percentage of sucrose (C₁₂H₂₂O₁₁) is 1.50%, we can first find the mass of the solute in grams and then find the number of moles. Calculate the mass of the solute: \(m_\mathrm{solute} = m_\mathrm{sol} \times \mathrm{mass \%} = (75.0 ~\mathrm{g}) (0.0150) = 1.125 ~\mathrm{g}\) Now, we need molar mass of sucrose: \(M_\mathrm{solute} = 12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342 ~\mathrm{g/mol}\). Finally, calculate the number of moles: \(n = \frac{m_\mathrm{solute}}{M_\mathrm{solute}} = \frac{1.125 ~\mathrm{g}}{342 ~\mathrm{g/mol}} = 0.00328 ~\mathrm{moles}\) So, there are 0.00328 moles of sucrose in the solution.