Final answer:
To covert obtain 198.8 mol of deuterium from natural hydrogen, which contains deuterium at a concentration of 0.01500%, approximately 11,927.99 kg (or about 12.8 kg) of water are required. So, the correct option is (c) 12.8 kg.
Step-by-step explanation:
The question relates to the amount of water needed to extract a given quantity of deuterium, based on its abundance in natural hydrogen.
To calculate this, we can use the given percentage of deuterium in hydrogen and the molar mass of water.
To find out how many kilograms of water are needed to obtain 198.8 mol of deuterium, we must first establish how many moles of hydrogen atoms this corresponds to, considering deuterium is a hydrogen isotope.
As the percentage of deuterium in natural hydrogen is given as 0.01500%, we can express this fraction as a decimal (0.00015) and use it to determine the total moles of hydrogen:
Total moles of hydrogen = 198.8 mol / 0.00015 = 1,325,333.33 mol of hydrogen
Each water molecule contains two hydrogen atoms, so the total moles of water would be half of the total moles of hydrogen molecules:
Total moles of water = 1,325,333.33 mol of hydrogen / 2 = 662,666.67 mol of water
The molar mass of water (H₂O) is approximately 18 g/mol, hence the total mass in kilograms would be:
Total mass of water = 662,666.67 mol * 18 g/mol / 1000 = 11927.99 kg of water
Therefore, the closest correct answer is (c) 12.8 kg.