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Question: \(150\,mL\) of \({C_2}{H_5}OH\) (density \( = \,\,0.78\,\,g\,\,m{L^{ - 1}}\) ) is diluted to one lit...

150 mL150\,mL of C2H5OH{C_2}{H_5}OH (density =  0.78  g  mL−1 = \,\,0.78\,\,g\,\,m{L^{ - 1}} ) is diluted to one litre by adding water; molarity of the solution is:
A) 2.542.54
B) 11.711.7
C) 2.992.99
D) 29.929.9

Explanation

Solution

To solve this question, we must first understand some basic concepts of Mole concept. Then we need to use the basic concepts and logic for calculation of the required ratio and then only we can conclude the correct answer.

Complete solution:
Before we move forward with the solution of this given question, let us first understand some basic concepts:
Molarity: is defined as the total number of moles of solute per litre of solution. The molality of a solution is dependent on the changes in physical properties of the system such as pressure and temperature as unlike mass, the volume of the system changes with the change in physical conditions of the system. Molarity is represented by M, which is termed as molar. One molar is the molarity of a solution where one gram of solute is dissolved in a litre of solution.
Step 1: In this step we will enlist all the given quantities:
Molecular weight of C2H5OH{C_2}{H_5}OH =  2×12+5×1+16+1  =  46 = \,\,2 \times 12 + 5 \times 1 + 16 + 1\,\, = \,\,46
Volume of C2H5OH{C_2}{H_5}OH =150  mL = 150\,\,mL
Density =  0.78  g/mL = \,\,0.78\,\,g/mL
Therefore, mass of C2H5OH{C_2}{H_5}OH = 0.78×150  =  117  g = \,0.78 \times 150\,\, = \,\,117\,\,g
Step 2: In this step we will calculate the Molarity:
Moles of C2H5OH{C_2}{H_5}OH   11746=  2.54\,\,\dfrac{{117}}{{46}} = \,\,2.54
Since the final volume of a solution is =  1  L = \,\,1\,\,L
Therefore, Molarity =  No.  of  moles  of  soluteVolume  of  solution  (in  L)=  2.541    2.54 = \,\,\dfrac{{No.\,\,of\,\,moles\,\,of\,\,solute}}{{Volume\,\,of\,\,solution\,\,(in\,\,L)}} = \,\,\dfrac{{2.54}}{1}\,\,\,\,2.54
Hence the required molarity is 2.542.54
So, clearly we can conclude that the correct answer is Option A.

Note: In thermodynamics the use of molar concentration is often not convenient because the volume of most solutions slightly depends on 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.