Question

Calculate the number of moles of solute present in each of the following aqueous solutions: (a) \(600 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{SrBr}_{2}\), (b) \(86.4 \mathrm{~g}\) of \(0.180 \mathrm{~m} \mathrm{KCl}\), (c) \(124.0 \mathrm{~g}\) of a solution that is \(6.45 \%\) glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) by mass.

Short answer

In summary, the number of moles of solute in each aqueous solution are: (a) \(0.150 \,\text{moles of}\, \mathrm{SrBr}_{2}\) (b) \(0.0155 \,\text{moles of}\, \mathrm{KCl}\) (c) \(0.0444 \,\text{moles of}\, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Step-by-step solution

Solve for (a) using volume and molarity

To find the number of moles of solute \(\left(\mathrm{SrBr}_{2}\right)\) in the solution, use the formula: Moles of solute = Volume of solution (in liters) × Molarity First, convert the volume to liters: \(600 \mathrm{~mL} = 0.600 \mathrm{~L}\) Now, multiply the volume in liters by the molarity of the solution: Moles of solute = \(0.600 \mathrm{~L} × 0.250 \mathrm{M}\) Moles of solute = \(0.150 \,\text{moles of}\, \mathrm{SrBr}_{2}\)

Solve for (b) using mass and molality

To find the number of moles of solute \(\left(\mathrm{KCl}\right)\) in the solution, use the formula: Moles of solute = Molality × Mass of solvent (in kilograms) First, convert the mass of solvent to kilograms: \(86.4 \mathrm{~g} = 0.0864 \mathrm{~kg}\) Now, multiply the mass of solvent in kilograms by the molality of the solution: Moles of solute = \(0.180 \mathrm{~m} × 0.0864 \mathrm{~kg}\) Moles of solute = \(0.0155 \,\text{moles of}\, \mathrm{KCl}\)

Solve for (c) using mass and mass percentage

To find the number of moles of solute \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in the solution, use the following steps: 1. Calculate the mass of glucose in the solution. 2. Calculate the moles of glucose using the molar mass of glucose. First, find the mass of glucose in the solution: Mass of glucose = \(6.45 \% × 124.0 \mathrm{~g}\) Mass of glucose = \(0.0645 × 124.0 \mathrm{~g} = 8.00 \mathrm{~g}\) Now, find the molar mass of glucose: Molar mass of glucose = \(6 × 12.01 \,\text{g/mol (C)} + 12 × 1.01 \,\text{g/mol (H)} + 6 × 16.00 \,\text{g/mol (O)} = 180.18 \,\text{g/mol}\) Finally, calculate the moles of glucose using the mass of glucose and its molar mass: Moles of solute = \(\frac{8.00 \mathrm{~g}}{180.18 \,\text{g/mol}}\) Moles of solute = \(0.0444 \,\text{moles of}\, \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) In summary, we have: (a) 0.150 moles of SrBr2 (b) 0.0155 moles of KCl (c) 0.0444 moles of C6H12O6