Question: What are the units that Molarity is expressed in? A.) Gram/litre B.) Moles/litre C.) Litre/mol...

Question

What are the units that Molarity is expressed in?
A.) Gram/litre
B.) Moles/litre
C.) Litre/mole
D.) Moles/1000 grams

Answer 1

Solution

Hint: The molarity of any chemical substance, be it a compound or an element, is given by the number of its moles present in one litre of solution. Analyse this definition and think of what could the units of molarity possibly be.

Complete step-by-step answer:
Let us observe the above definition and come up with an equation to express Molarity, so as to help us understand what it's possible units could be.

$Molarity(M)=\text{ }\dfrac{\text{Number of moles of solute (m)}}{\text{Volume of solution (V)}}$

Remember that the SI Unit for the amount of substance is moles, or mol in short.
Also remember that the widely accepted and the most commonly known unit for volume is Litre, or l in short.
With this information, we can hereby conclude that the unit of Molarity is given by:

$\text{Unit of }Molarity(M)=\text{ }\dfrac{mol}{litre}$

Which can also be expressed as $mol\text{ }{{l}^{-1}}$
Let us take an example to fully understand the above explanation. Let us try and calculate the molarity of 117 g of common salt (NaCl) dissolved in 1.5l of water.
Calculating the number of moles of NaCl present in solution:
$\begin{aligned} & Moles(m)=\dfrac{Weight}{Molar\text{ }Mass} \\\ & \Rightarrow m=\dfrac{117}{58.5}=2 \\\ \end{aligned}$
Given that the volume of solution in this instance is 1.5 litre, let us now apply the obtained data into this formula:
$Molarity(M)=\text{ }\dfrac{\text{Number of moles of solute (m)}}{\text{Volume of solution (V)}}$
$M=\dfrac{2\text{ moles}}{1.5\text{ l}}\approx 1.33\text{ mol }{{\text{l}}^{\text{-1}}}$
Now, not only have established what the unit of molarity is conceptually, we have also used an example to explain the same.
Therefore, the answer to this question is b)

NOTE: Be very careful about the definition of molarity while solving that question and remember that it is the number of moles present in 1l of solution and not the weight of the substance present for the same.