Calculate the number of moles of solute present in each of the following solutions: (a) \(255 \mathrm{~mL}\) of \(1.50 \mathrm{M} \mathrm{HNO}_{3}(a q)\) (b) \(50.0 \mathrm{mg}\) of an aqueous solution that is $1.50 \mathrm{~m} \mathrm{NaCl}$, (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.

To find the moles of HNO3, we are given its molarity and the volume of the solution. We can use the formula: Moles of solute = Molarity × Volume (in L) We are given Molarity=1.50 M and Volume= 255 mL. First, convert the volume to liters: Volume (L) = 255 mL × (1 L / 1000 mL) = 0.255 L Now, calculate the moles of HNO3: Moles of HNO3 = (1.50 M) × (0.255 L) = 0.3825 mol

To find the moles of NaCl, we are given the mass of the solution and its molality. We can use the formula: Moles of solute = (Molality × Mass of solvent)/(Molar mass of solute) We are given Molality = 1.50 m, and the Mass of the solution = 50.0 mg and the molar mass of NaCl = 58.44 g/mol. First, convert the mass of the solution to grams: Mass (g) = 50.0 mg × (1 g / 1000 mg) = 0.0500 g Now, calculate the mass of the solvent: Mass of solvent = Mass of solution - Mass of solute Since the molality is given, we can write the equation: Mass of Solute = Molality × Mass of solvent First, we need to find the mass of solvent. Rearranging, Mass of solvent = Mass of Solute / Molality We can denote the mass of solvent by x and the mass of solute by y: 0.0500 g = x + y y = (1.50 m) × x y = (1.50) × (0.0500 g - y) Solving for y, we get y = 0.023148 g Now, find the moles of NaCl: Moles of NaCl = mass of solute / Molar mass of solute Moles of NaCl = 0.023148 g / 58.44 g/mol = 3.96 × 10^{-4} mol

To find the moles of sucrose, we are given the mass percentage, and we can first find mass of the solute and then we can find moles using the molar mass. We are given mass percentage = 1.50 % and the Mass of the solution = 75.0 g and the molar mass of sucrose = 342.3 g/mol. First, calculate the mass of sucrose: Mass of sucrose = (1.50 %) × (75.0 g) = 1.125 g Now, find the moles of sucrose: Moles of sucrose = mass of solute / Molar mass of solute Moles of sucrose = 1.125 g / 342.3 g/mol = 3.29 × 10^{-3} mol Now we have the number of moles of solute present in each solution: (a) 0.3825 mol of HNO3 (b) 3.96 × 10^{-4} mol of NaCl (c) 3.29 × 10^{-3} mol of sucrose