Question

What is molarity of a solution of $HCl$ which contains 49 % by weight of solute and whose specific gravity is 1.41?
A) 15.25
B) 16.75
C) 18.92
D) 20.08

Answer 1

Hint: The answer to this question is based on the calculation of molarity of the solution which is given by the formula according to the definition as $Molarity=\dfrac{{{n}_{solute}}}{{{V}_{solution(litres)}}}$ and this value gives the correct answer.

Complete Solution :
In the lower classes of physical chemistry, we have studied the basic measurement parameters of the solutions like molarity, molality, normality and also mole fraction of the solutions.
- Now, let us see in detail how molarity is calculated according to the data given to us.
In the given data, we have the values, 49 % by weight of solute that is the $HCl$ which is nothing but 49 % by weight by volume of $HCl$ solution.
Also, the specific gravity of solution is 1.41.
Therefore, from this data we can calculate weight of $HCl$ solution in 1 ml of the solution as follows,
Weight of $HCl$ in 1 ml solution = specific gravity $\times $ 49 %
\[\Rightarrow {{w}_{HCl}}=1.41\times \dfrac{49}{100} = 0.6909g/ml\]
Therefore, we have 0.6909 g of $HCl$ in 1 ml of the solution.

- Now, according to the definition of molarity that is the ratio of number of moles of solute to that of volume of the solution in litres, we can write the formula as
$Molarity=\dfrac{{{n}_{solute}}}{{{V}_{solution(litres)}}}$
Number of moles of solute that is $HCl$ is total weight divided by the molecular weight that is ${{n}_{HCl}}=\dfrac{w}{m}=\dfrac{0.6909}{36.4} = 0.0189g/mol$
Thus, substituting the value of the weight for one litre of solution and also the number of moles of hydrochloric acid found in the molarity formula, we have
\[Molarity = \dfrac{06909}{0.0189} = 18.92M\]
So, the correct answer is “Option C”.

Note: Note that specific gravity which is also called as relative density is nothing but the ratio of density of the substance to the density of the given reference material and therefore it does not contain any units and when the data is given in terms of only density then calculate the values accordingly and do not be confused.