The mass defect for a lithium-6 nucleus is \(-0.03434 \mathrm{~g} / \mathrm{mol}\). Calculate the atomic mass of \({ }^{6} \mathrm{Li}\)

Since lithium-6 has a mass number of 6 and a known atomic number of 3, we can determine that there are 3 protons and 3 neutrons in its nucleus.

We know that there are 3 protons and 3 neutrons in the nucleus of lithium-6, and also 3 electrons orbiting it. To find the summed mass, we'll need the mass of each particle: - Mass of a proton: \(1.0073 \mathrm{~amu}\) - Mass of a neutron: \(1.0087 \mathrm{~amu}\) - Mass of an electron: \(5.4858 \times 10^{-4} \mathrm{~amu}\) Now calculate the combined mass: \[ (\text{total_protons_mass}) = 3 * 1.0073 \mathrm{~amu} = 3.0219 \mathrm{~amu} \] \[ (\text{total_neutrons_mass}) = 3 * 1.0087 \mathrm{~amu} = 3.0261 \mathrm{~amu} \] \[ (\text{total_electrons_mass}) = 3 * (5.4858 \times 10^{-4} \mathrm{~amu}) = 1.6457 \times 10^{-3} \mathrm{~amu} \] Sum up the masses: \[ (\text{total_mass}) = (\text{total_protons_mass}) + (\text{total_neutrons_mass}) + (\text{total_electrons_mass}) = 3.0219 + 3.0261 + 1.6457 \times 10^{-3} \mathrm{~amu} \]

We are given the mass defect in grams per mole, but our calculations are in atomic mass units. To convert, we will use the conversion factor \(1 \mathrm{~amu} = \frac{1}{N_A} \mathrm{~g\, mol}^{-1}\), where \(N_A\) is Avogadro's Number (\(6.0221 \times 10^{23}\) atoms/mol): \[ \text{mass_defect_amu} = -0.03434\,\frac{\mathrm{g} }{\mathrm{mol} } \times \frac{1\,\mathrm{amu}}{1 / 6.0221 \times 10^{23}\, \mathrm{atoms/mol} } \]

We can now calculate the atomic mass of lithium-6 using the total mass calculated in step 2 and the converted mass defect from step 3. The atomic mass can be found by subtracting the mass defect in amu from the total mass: \[ \text{atomic_mass_Li6} = (\text{total_mass}) - (\text{mass_defect_amu}) \] Calculate the atomic mass of lithium-6 based on the above expression.