Determine how many grams of each of the following solutes would be needed to make \(2.50 \times 10^{2} \mathrm{~mL}\) of a \(0.100 M\) solution: (a) cesium iodide (CsI), (b) sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right),\) (c) sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3}\right),\) (d) potassium dichromate \(\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)\) (e) potassium permanganate \(\left(\mathrm{KMnO}_{4}\right)\).

In order to use the formula of molar concentration, we need the volume in liters. \(1 \mathrm{~mL}\) is equal to \(0.001 \mathrm{~L}\), so \(2.50 \times 10^{2} \mathrm{~mL}\) of solution is equal to \(2.50 \times 10^{2} \times 0.001 = 0.250 \mathrm{~L}\)

Since we know the molar concentration and the volume of the solution, we can find the amount of solute in moles using the formula \(n = MV\). So, we have \(n = 0.100 M \times 0.250 L = 0.0250 \mathrm{~moles}\) of solute needed for each solution.

We now find the mass of each solute by multiplying the number of moles by the molar mass of the respective solute: \n(a) For CsI, the molar mass is \(132.91 g/mol + 126.90 g/mol = 259.81 g/mol\). Therefore, the mass is \(0.0250 mol \times 259.81 g/mol = 6.50 g\). \n(b) For \(\mathrm{H}_{2} \mathrm{SO}_{4}\), the molar mass is \(2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol\). Therefore, the mass is \(0.0250 mol \times 98.09 g/mol = 2.45 g\). \n(c) For \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), the molar mass is \(2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol\). Therefore, the mass is \(0.0250 mol \times 105.99 g/mol = 2.65 g\). \n(d) For \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\), the molar mass is \(2(39.10 g/mol) + 2(52.00 g/mol) + 7(16.00 g/mol) = 294.20 g/mol\). Therefore, the mass is \(0.0250 mol \times 294.20 g/mol = 7.35 g\). \n(e) For \(\mathrm{KMnO}_{4}\), the molar mass is \(39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol) = 158.04 g/mol\). Therefore, the mass is \(0.0250 mol \times 158.04 g/mol = 3.95 g\).