# How many kilograms of waters are needed to obtain the

How many kilograms of waters are needed to obtain the 198.8 mol of deuterium, assuming that deiterium is 0.01500% of natural hydrogen?
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Haylie Cherry
Given: Number of moles of deuterium is 198.8 mol.
The percentage of deuterium by number of natural hydrogen is 0.01500%.
To determine: The water needed in kilograms.
The molecular weight of water ${H}_{2}O$ is 18 g.
Percentage of deuterium present in hydrogen is 0.01500% by number, that is for 100 mole of Hydrogen, there are 0.0150 mole of deuterium.
One mole of water has two mole of hydrogen and one mole of oxygen.
The required amount of deuterium is 198.8 mol.
The equation to calculate the mass of water is
$m=n×\frac{{N}_{H}}{N}×\frac{{N}_{w}}{{N}_{{H}_{w}}}$
where m is the required mass of water.
n is the required amount of deuterium that is 198.8 mol.
${N}_{H}$ is the number of moles of hydrogen that are initially present that is 100 mol.
N is the number of moles of deuterium that is 0.0150 mol.
${N}_{w}$ is the mass of water that is 18 g.
${N}_{{H}_{w}}$ is the number of moles of hydrogen in water, that is 2 mol.
Substitute these values in equation:

Answer: The amount of water is 11928 kg.