Final answer:
The mass percent of water in the hydrate LiNO3·3H2O is calculated by finding the mass of water in the hydrate, dividing it by the molar mass of the hydrate, then multiplying by 100%. The result is approximately 43.94%, with the closest provided answer being 42.37% (D).
Step-by-step explanation:
To calculate the mass percent of water in the hydrate LiNO₃·3H₂O, you must find the mass of water in one mole of the hydrate, then divide that by the molar mass of the entire hydrate, and multiply by 100%. First, calculate the molar mass of LiNO₃, which is the sum of the molar masses of Li (6.94 g/mol), N (14.01 g/mol), and 3 O (3 × 16.00 g/mol). This equals 68.94 g/mol for LiNO₃.
The molar mass of water (H₂O) is 18.015 g/mol. Since there are three moles of H₂O, the total mass of water is 3 × 18.015 g/mol = 54.045 g/mol. Now, add this to the molar mass of LiNO₃ to get the molar mass of the entire hydrate: 68.94 g/mol + 54.045 g/mol = 122.985 g/mol.
Finally, calculate the mass percent of water: (54.045 g/mol / 122.985 g/mol) × 100% = 43.94%. Therefore, the closest answer to the choices given would be 42.37% (D), which might be an approximate value provided in the question.