Question

Question: Heavy water, \({{\rm{D}}_{\rm{2}}}{\rm{O}}\) (molar mass\({\rm{ = 20}}{\rm{.03 gmo}}{{\rm{l}}^{{\rm{ - 1}}}}\)), can be separated from ordinary water, \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) (molar mass\({\rm{ = 18}}{\rm{.01}}\)), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) and \({{\rm{D}}_{\rm{2}}}{\rm{O}}\).

Short answer

The relative rates of diffusion of \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) and \({{\rm{D}}_{\rm{2}}}{\rm{O}}\) is \({\rm{1}}{\rm{.054}}{\rm{.}}\)

Step-by-step solution

Concept Introduction

The Graham's law states that a gas's rate of diffusion or effusion is inversely related to its square root of molecular weight.

Relative Rates of Diffusion

Graham's finding relates the rate and the molar mass of a gas. If two gases \(A\) and \(B\) are at the same temperature and pressure, then the relative rates of the diffusion for \({{\rm{H}}_{\rm{2}}}{\rm{O}}\) and \({{\rm{D}}_{\rm{2}}}{\rm{O}}\) will be:

\(\begin{array}{l}\frac{{{\rm{ rate of diffusion of }}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{ rate of diffusion of }}{{\rm{D}}_{\rm{2}}}{\rm{O}}}}{\rm{ = }}\frac{{\sqrt {{{\rm{M}}_{{{\rm{D}}_{\rm{2}}}{\rm{O}}}}} }}{{\sqrt {{{\rm{M}}_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}} }},\;where\;M\;is\;the\;molar\;mass.\\{\rm{ = }}\sqrt {\frac{{{\rm{20}}{\rm{.028}}}}{{{\rm{18}}{\rm{.015}}}}} \\{\rm{ = }}\sqrt {{\rm{1}}{\rm{.1117}}} \\{\rm{ = 1}}{\rm{.054}}{\rm{.}}\end{array}\)

Therefore, the result obtained is \({\rm{1}}{\rm{.054}}{\rm{.}}\)