Question

How many kilograms of water are needed to obtain the \(198.8{\rm{ }}mol\) of deuterium, assuming that deuterium is \(0.01500\% \) (by number) of natural hydrogen?

Short answer

To obtain \(198.8{\rm{ }}mol\) of deuterium, assuming that deuterium is \(0.01500\% \) of natural hydrogen, the mass of water needed is \(m = 11928\;kg\).

Step-by-step solution

Concept Introduction

The power kept in the centre of an atom is known as nuclear energy. All stuff in the cosmos is made up of tiny particles called atoms. The centre of an atom's nucleus is typically where most of its mass is concentrated. Neutrons and protons are the two subatomic particles that make up the nucleus. Atomic bonds that hold them together carry a lot of energy.

Information Provided

• Amount of natural hydrogen in deuterium is:\(0.01500\% \).
• Amount of deuterium is: \(198.8{\rm{ }}mol\).

Amount of water

Assuming that deuterium is \(0.01500\% \) of hydrogen. The molar mass for hydrogen is \(1{\rm{ }}g/mol\), and oxygen is \(16{\rm{ }}g/mol\), which means water will have \(18{\rm{ }}g/mol\).

There is a need of \(198.8{\rm{ }}mol\) pure deuterium. Since in hydrogen, there is only \(0.01500\% \) of deuterium it means, that if \(100{\rm{ }}g/mol\) of hydrogen is there, then \(0.0150{\rm{ }}g/mol\) of deuterium will be present, so to get the needed amount –

\(m = {\rm{ }}deuterium{\rm{ }} \cdot \frac{{{\rm{ }}hydrogen{\rm{ }}}}{{{\rm{ }}deuterium{\rm{ }}in{\rm{ }}hydrogen{\rm{ }}}} \cdot \frac{{{\rm{ }}water{\rm{ }}}}{{{\rm{ }}hydrogen{\rm{ }}in{\rm{ }}water{\rm{ }}}}\)

Plugging in the values –

\(m = 198.8\;mol \cdot \frac{{100\;mol}}{{0.0150\;mol}} \cdot \frac{{18\;g}}{{2\;mol}}\)

Hence, the result is obtained as –

\(\begin{array}{c}m = 198.8\;mol \cdot \frac{{100\;mol}}{{0.0150\;mol}} \cdot \frac{{18\;g}}{{2\;mol}}\\m = 1192800\;g\\ = 11928\;kg\end{array}\)

Therefore, the value for mass of water is obtained as \(m = 11928\;kg\).