91Ó°ÊÓ

First, we need to convert the volume of the solution from mL to L: \(500.0\,\text{mL} \times \frac{1\,\text{L}}{1000\,\text{mL}} = 0.500\,\text{L}\) Now, we can use the formula for molarity to find the number of moles of KBr: Molarity = \(\frac{\text{moles of solute}}{\text{liters of solution}}\) \(0.250\,M = \frac{\text{moles of KBr}}{0.500\,\text{L}}\) Solve for moles of KBr: Moles of KBr = \(0.250\,M \times 0.500\,\text{L} = 0.125\,\text{mol}\) Next, we'll need the molar mass of KBr, which is: \(39.10\,\frac{\text{g}}{\text{mol}}\) (K) + \(79.90\,\frac{\text{g}}{\text{mol}}\) (Br) = \(119.00\,\frac{\text{g}}{\text{mol}}\) Finally, we can convert moles of KBr to grams: Grams of KBr = \((0.125\,\text{mol}) \times (119.00\,\frac{\text{g}}{\text{mol}}) = 14.88\,\text{g}\)

First, we need to convert the volume of the solution from mL to L: \(25.0\,\text{mL} \times \frac{1\,\text{L}}{1000\,\text{mL}} = 0.025\,\text{L}\) Now, use the formula for molarity to find the number of moles of NaNO3: Molarity = \(\frac{\text{moles of solute}}{\text{liters of solution}}\) \(0.200\,M = \frac{\text{moles of NaNO3}}{0.025\,\text{L}}\) Solve for moles of NaNO3: Moles of NaNO3 = \(0.200\,M \times 0.025\,\text{L} = 0.005\,\text{mol}\) Next, we'll need the molar mass of NaNO3, which is: \(22.99\,\frac{\text{g}}{\text{mol}}\) (Na) + \(14.01\,\frac{\text{g}}{\text{mol}}\) (N) + \(3\times 16.00\,\frac{\text{g}}{\text{mol}}\) (O) = \(85.00\,\frac{\text{g}}{\text{mol}}\) Finally, we can convert moles of NaNO3 to grams: Grams of NaNO3 = \((0.005\,\text{mol}) \times (85.00\,\frac{\text{g}}{\text{mol}}) = 0.425\,\text{g}\)

First, we need to convert the volume of the solution from mL to L: \(100.0\,\text{mL} \times \frac{1\,\text{L}}{1000\,\text{mL}} = 0.100\,\text{L}\) Now, use the formula for molarity to find the number of moles of CH3OH: Molarity = \(\frac{\text{moles of solute}}{\text{liters of solution}}\) \(0.375\,M = \frac{\text{moles of CH3OH}}{0.100\,\text{L}}\) Solve for moles of CH3OH: Moles of CH3OH = \(0.375\,M \times 0.100\,\text{L} = 0.0375\,\text{mol}\) Next, we'll need the molar mass of CH3OH, which is: \(12.01\,\frac{\text{g}}{\text{mol}}\) (C) + \(3\times 1.01\,\frac{\text{g}}{\text{mol}}\) (H) + \(16.00\,\frac{\text{g}}{\text{mol}}\) (O) + \(1.01\,\frac{\text{g}}{\text{mol}}\) (H) = \(32.04\,\frac{\text{g}}{\text{mol}}\) Finally, we can convert moles of CH3OH to grams: Grams of CH3OH = \((0.0375\,\text{mol}) \times (32.04\,\frac{\text{g}}{\text{mol}}) = 1.201\,\text{g}\)