Boron consists of two isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B}\). Chlorine also has two isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\). How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?

First, we need to list all possible combinations of the Boron and Chlorine isotopes in a BCl₃ molecule. Since there are two isotopes of Boron and two isotopes of Chlorine, we have four possible combinations: 1. \({ }^{10} \mathrm{B}\) and \({ }^{35} \mathrm{Cl}\) 2. \({ }^{10} \mathrm{B}\) and \({ }^{37} \mathrm{Cl}\) 3. \({ }^{11} \mathrm{B}\) and \({ }^{35} \mathrm{Cl}\) 4. \({ }^{11} \mathrm{B}\) and \({ }^{37} \mathrm{Cl}\)

Now, we will calculate the mass of the BCl₃ molecule for each combination of isotopes. For a given combination, the mass of BCl₃ will be the sum of the mass of the Boron isotope and the mass of three Chlorine isotopes: 1. For combination 1 ( \({ }^{10} \mathrm{B}\)+3×\({ }^{35} \mathrm{Cl}\)): mass = 10 + 3 × 35 = 10 + 105 = 115 2. For combination 2 ( \({ }^{10} \mathrm{B}\)+2×\({ }^{35} \mathrm{Cl}\)+\({ }^{37} \mathrm{Cl}\)): mass = 10 + 2 × 35 + 37 = 10 + 70 + 37 = 117 3. For combination 3 ( \({ }^{11} \mathrm{B}\)+3×\({ }^{35} \mathrm{Cl}\)): mass = 11 + 3 × 35 = 11 + 105 = 116 4. For combination 4 ( \({ }^{11} \mathrm{B}\)+2×\({ }^{35} \mathrm{Cl}\)+\({ }^{37} \mathrm{Cl}\)): mass = 11 + 2 × 35 + 37 = 11 + 70 + 37 = 118

Now we have the mass of each combination of isotopes in BCl₃: 1. Combination 1: 115 2. Combination 2: 117 3. Combination 3: 116 4. Combination 4: 118 We can see that all four combinations have different masses. Therefore, there will be four unique peaks in the mass spectrum of BCl₃. The approximate mass of each peak is as follows: 1. Peak 1: 115 2. Peak 2: 116 3. Peak 3: 117 4. Peak 4: 118