Momple Choice 1 point Calculate the percentage of water in the following hydrates: beryllium nitrate trihydrate, Be(NO_(3))_(2)cdot 3H_(2)O 45.66% water 15.45% water 67.89% water 28.90% water

To calculate the percentage of water in beryllium nitrate trihydrate, we need to find the molar mass of the compound and the molar mass of the water molecules.<br /><br />The molar mass of beryllium nitrate trihydrate is calculated as follows:<br />Molar mass of Be = 9.01 g/mol<br />Molar mass of N = 14.01 g/mol<br />Molar mass of O = 16.00 g/mol<br />Molar mass of H = 1.01 g/mol<br /><br />Molar mass of Be(NO3)2 = (9.01 g/mol) + 2 * [(14.01 g/mol) + 3 * (16.00 g/mol)]<br />Molar mass of Be(NO3)2 = 9.01 g/mol + 2 * (14.01 g/mol + 48.00 g/mol)<br />Molar mass of Be(NO3)2 = 9.01 g/mol + 2 * 62.01 g/mol<br />Molar mass of Be(NO3)2 = 9.01 g/mol + 124.02 g/mol<br />Molar mass of Be(NO3)2 = 133.03 g/mol<br /><br />The molar mass of the water molecules is calculated as follows:<br />Molar mass of H2O = (2 * 1.01 g/mol) + 16.00 g/mol<br />Molar mass of H2O = 2.02 g/mol + 16.00 g/mol<br />Molar mass of H2O = 18.02 g/mol<br /><br />The molar mass of the water molecules in the compound is calculated as follows:<br />Molar mass of 3H2O = 3 * 18.02 g/mol<br />Molar mass of 3H2O = 54.06 g/mol<br /><br />The percentage of water in the compound is calculated as follows:<br />Percentage of water = (Molar mass of 3H2O / Molar mass of Be(NO3)2) * 100<br />Percentage of water = (54.06 g/mol / 133.03 g/mol) * 100<br />Percentage of water = 40.66%<br /><br />Therefore, the correct answer is $40.66 water$.