Calculate the molar mass of \( \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \) \( 199.3 \mathrm{~g} / \mathrm{mol} \) \( 261.3 \mathrm{~g} / \mathrm{mol} \) \( 398.6 \mathrm{~g} / \mathrm{mol} \) \( 323.3 \mathrm{~g} / \mathrm{mol} \) \( 247.3 \mathrm{~g} / \mathrm{mol} \)

To calculate the molar mass of \( \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \), we first identify the individual atomic masses from the periodic table: - Barium (Ba): approximately 137.3 g/mol - Nitrogen (N): approximately 14.0 g/mol (and there are 2 N atoms in the formula) - Oxygen (O): approximately 16.0 g/mol (and there are 6 O atoms in the formula, since there are two nitrate ions) Now we can sum up the mass contributions: - Molar mass of Ba: \( 137.3 \, \text{g/mol} \) - Molar mass of Nitrogen in 2 nitrate: \( 2 \times 14.0 \, \text{g/mol} = 28.0 \, \text{g/mol} \) - Molar mass of Oxygen in 6 oxygen: \( 6 \times 16.0 \, \text{g/mol} = 96.0 \, \text{g/mol} \) Adding these together: \[ 137.3 \, \text{g/mol} + 28.0 \, \text{g/mol} + 96.0 \, \text{g/mol} = 261.3 \, \text{g/mol} \] So, the molar mass of \( \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} \) is \( 261.3 \, \text{g/mol} \). And that’s the answer!