Question

The molarity of heavy water ${ D }_{ 2 }O$ is :[assume density of ${ D }_{ 2 }O$ is 1g/ml]
A. 55.55 M
B. 50.78 M
C. 48 M
D. 50 M

Answer 1

Hint: To calculate its molarity we need two constants. We already know its density and we can find its molar mass, simply from its molecular formula.

Complete step by step answer:

Heavy water, ${ D }_{ 2 }O$ is what you get when you replace the hydrogen-1 isotopes, that are usually present in the vast majority of water molecules, with deuterium or hydrogen-2 isotopes.
The difference between hydrogen-1 and deuterium is that the latter contains one proton and one neutron inside its nucleus, as opposed to the former which only contains a proton.
We already know density(d) of ${ D }_{ 2 }O$ = 1g/ml
Molecular weight (M) of ${ D }_{ 2 }O$ = 2(2.0138) + 16 = 20.0276 g/mol
Let's assume we have 1L of heavy water.
Molarity - It is the concentration of a solution expressed as the number of moles of solute per liter of solution.
Hence, Molarity is given by the formula (in terms of density)
$\quad =\quad \dfrac { Density\quad \times \quad 1000 }{ Molecular\quad weight\quad \times \quad Volume }$

Now we will insert the values for each quantity i.e. density (in g/ml), molecular weight (in g/mol) and volume (in liters) in the above formula. The value of Molarity (M) will be equal to,
                    $=\quad \dfrac { 1\quad \times \quad 1000 }{ 20.0276\quad \times \quad 1 }$
                    $=\quad \dfrac { 1000 }{ 20.0276 }$
Molarity (M) = 49.93M
We can see this is approximately equal to 50M.

Therefore, the correct answer for this question is option D.

Note: The presence of deuterium gives the water different nuclear properties, and the increase of mass gives it slightly different physical and chemical properties when compared to normal water.
${ D }_{ 2 }O$ is used in Nuclear Magnetic Resonance (NMR) spectroscopy, in organic chemistry, Fourier Transform Infrared (FTIR) spectroscopy, and in some types of nuclear reactors as a moderator to slow down the velocity of neutrons.