How many milligrams \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be present in \(50.0 \mathrm{L}\) of a solution containing \(2.35 \mathrm{ppm} \mathrm{Ca} ?\) [Hint: See also Exercise 103 .]

To convert the concentration from ppm to milligrams, we need to understand that 1 ppm is equivalent to 1 milligram per liter (\(mg/L\)). Therefore, the concentration of Calcium in the solution is 2.35 \(mg/L\).

Multiply the volume of the solution by the concentration to find the total weight of Calcium. This is done because the concentration is the amount of a substance per unit volume. So, the total weight of Calcium is equal to 2.35 \(mg/L\) * 50.0 \(L\) = \(117.5 mg\).

To calculate the mass of \(Ca(NO_3)_2\), it is necessary to understand that the weight of Calcium is part of the total weight of \(Ca(NO_3)_2\). So, we need to find the molar mass of Calcium and \(Ca(NO_3)_2\). The atomic masses are approximately: Calcium = 40, Nitrogen = 14, Oxygen = 16. Therefore, the molar mass of \(Ca(NO_3)_2\) is approximately \(40 + 2*(14 + 3*16) = 164 g/mol\). Given that we are dealing with milligrams in this question, it is more convenient to express the molar mass in \(mg\), therefore, the molar mass is \(164,000 mg/mol\). To find the mass of \(Ca(NO_3)_2\) that relates to 117.5 \(mg\) of Calcium, we use the formula: \((117.5 mg \, Ca \times 164,000 mg \, Ca(NO_3)_2)/(40,000 mg \, Ca) = 480.25 mg \, Ca(NO_3)_2\).