Question

# The percentage of oxygen in heavy water is$A.60 \\ B.50 \\ C.80 \\ D.20 \\$

Hint: We must know that the heavy water is a form of water that contains deuterium instead of hydrogen. Therefore, the molecular formula of heavy water is ${D_2}O$. We have to find the percentage of oxygen present in the heavy water.
Let’s start by discussing a little about the heavy water; heavy water is a form of water that contains deuterium instead of hydrogen. Deuterium is an isotope of hydrogen which carries 1 proton and 1 neutron in its nucleus whereas the hydrogen carries only 1 proton in its nucleus. It is heavier than water hence is known as heavy water. Due to the presence of deuterium the water is having different nuclear properties and an increase in mass results in different physical and chemical properties as compared with normal water. It is not radioactive in nature and in its pure form is having $11\%$ more mass than normal water.
Now coming back to question, we know that the molecular formula for heavy water will be ${D_2}O$. The atomic mass of Deuterium and oxygen is 2 and 16 respectively so, the molecular mass of the heavy water will be
Molecular mass of heavy water $= {\text{ }}2{\text{ }}X{\text{ }}2{\text{ }} + {\text{ }}16{\text{ }} = {\text{ }}20$
Percentage of oxygen in heavy water = $\dfrac{{{\text{Mass of oxygen in }}{{\text{D}}_{\text{2}}}{\text{0}}}}{{{\text{Molecular Mass of }}{{\text{D}}_{\text{2}}}{\text{O}}}} = \dfrac{{16}}{{20}} \times 100 = 80\%$
We must know the uses of heavy water in different applications. Therefore, the major role of heavy water is seen in nuclear reactors, where they act as nuclear moderators, which slows down the neutrons so that they can act with fissile uranium-235 rather than uranium 238, which captures the neutrons without fissioning. It is preferred in place of light water $\left( {{H_2}O} \right)$ because light water absorbs more neutrons than ${D_2}O$.