Question

If you mix 3 Litres of 0.5M NaCl with 9 litres of 0.2777M NaCl, what will be the concentration of the final solution, be assuming that the volumes are additive?
A.0.33M
B.0.39M
C.0.5777M
D.0.6777M
E.None of these

Answer 1

The two solutions are given and the concentration term used over here is Molarity and this question could be answered by using the molarity equation i.e. ${M_1}{V_1} = {M_2}{V_2}$ and in this solution and mixture could be compared to answer the question.

Complete step by step answer:
-In the given question, there are two solutions provided which are then finally mixed to form a new solution. We have to find out the concentration (molarity in this case) of the resultant solution.
-Here the concentration of the solutions is expressed in Molarity which means the number of moles of solute present per litre of the solution and the solution are of $NaCl$ which is also called as common salt.
-So, there is 3 litre of 0.5M NaCl and let us consider it as solution 1 and there is another solution which is 9 litre of 0.277M NaCl and let us consider it as solution 2.
-So, the molarity of the resultant solution is given by the equation (molarity of mixture of solution): -
${M_{mixture}} = \dfrac{{({M_1}{V_1} + {M_2}{V_2})}}{{{V_{total}}}}$ and this equation came by: - ${M_1}{V_1} + {\text{ }}{M_2}{V_2} = {\text{ }}{M_{mix}}{V_{mix}}$
-It is also given in the question that the volumes are additive and molarity of mixture is calculated by this formula. So, putting all the respective values of solution 1 and solution 2, we get: -
${M_{mixture}} = \dfrac{{(0.5 \times 3 + 0.277 \times 9)}}{{3 + 9}} = \dfrac{{1.5 + 2.5}}{{12}} = 0.33$
So, the concentration i.e. molarity of the resultant mixture is 0.33M

Therefore, the correct answer is A i.e. 0.33M.

Note:
The concentration term used in this concept is Molarity and there are different more concentration terms like Molality, Normality, Mole dfraction. Molarity is always preferred over Molality as it is independent of temperature.