Question

$ 1M\;HCl $ and $ 2M\;H\;Cl $ are mixed I volume ratio $ 4:1 $ . What is the final molarity of $ HCl $ solution?
(A) $ 1.5 $
(B) $ 1 $
(C) $ 1.2 $
(D) $ 1.8 $

Answer 1

Hint: To find the total molarity of the given solution first find the number of moles of the solution in each case with the help of molarity and the volume ratio given. Then simply derive the formula for the molarity when they are mixed in the fixed ratio.

Complete answer:
Molarity is defined as the number of moles of solutes present in one litre of solution. It is denoted by $ M $ . Thus molarity is-
 $ M = \dfrac{{No.{\text{ }}of{\text{ }}moles{\text{ }}of{\text{ }}solute}}{{{\text{Volume }}of{\text{ }}solution{\text{ }}in{\text{ Litre}}}} $
 $ M = \dfrac{n}{V} $
Where $ n $ is the number of moles
The SI unit of molarity is $ mol/L $ .
Molarity also known as molar concentration is not often used in thermodynamics because the volume of most solutions slightly depends on the temperature due to thermal expansion. This problem is usually resolved by introducing temperature correction factors, or by using a temperature independent measure of concentration such as molality.
The molar concentration depends on the variation of the volume of the solution due to thermal expansion.
We will assume the volume of $ 2M $ to be $ xL $ and the volume of $ 1M $ of solution to be $ 4xL $ as the ratio of volume given is $ 4:1 $
To calculate the number of moles, it is the ratio of the given mass of the substance or compound present in the sample to the weight of the substance by its molar mass. To find the number of moles of $ 1M $ and $ 2M $ solution of $ H\;Cl $ we will multiply its molarity to its volume. This can be given by the formula-
No. of moles of solution= $ molarity \times volume $
Therefore
No of moles of $ 1M $ of $ H\;Cl = {M_1}{V_1} $
No of moles of $ 2M $ of $ H\;Cl = {M_2}{V_2} $
Putting together all these into one formula we will get
Molarity = $ \dfrac{{{M_1}{V_1} + {M_2}{V_2}}}{{{V_1} + {V_2}}} $
Where $ {M_1} = 1M $ , $ {V_1} = 4x $ , $ {M_2} = 2M $ , $ {V_2} = x $
Putting the above value in the given formula
 $ = \dfrac{{1 \times 4x + 2 \times x}}{{4x + x}} $
Canceling the variable from the numerator and the denominator
 $ = \dfrac{{4 + 2}}{5} $
 $ = \dfrac{6}{5} = 1.2M $
Therefore our answer is option C.

Note:
Molarity and molality are two different ways of expressing concentration of solutions. Molarity is the number of moles of the solute in one litre of solution whereas molality is expressed as the number of moles in one kilogram of solvent.