Question

Molarity equation of a mixture of solutions of same substance is given by:
(a) ${ M }_{ 1 }+{ V }_{ 1 }\times { M }_{ 2 }+{ V }_{ 2 }\times { M }_{ 3 }+{ V }_{ 3 }+...={ M }_{ 1 }+{ M }_{ 2 }+{ M }_{ 3 }$
(b) ${ M }_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }+{ M }_{ 3 }{ V }_{ 3 }+...={ M }({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 })$
(c) $\cfrac { { M }_{ 1 } }{ { V }_{ 1 } } +\cfrac { { M }_{ 2 } }{ { V }_{ 2 } } +\cfrac { { M }_{ 3 } }{ { V }_{ 3 } } +...=M(\cfrac { 1 }{ { V }_{ 1 } } +\cfrac { 1 }{ { V }_{ 2 } } +\cfrac { 1 }{ { V }_{ 3 } } )$
(d) $\cfrac { { M }_{ 1 } }{ { V }_{ 1 } } \times \cfrac { { M }_{ 2 } }{ { V }_{ 2 } } \times \cfrac { { M }_{ 3 } }{ { V }_{ 3 } } +...=M(\cfrac { 1 }{ { V }_{ 1 } } \times \cfrac { 1 }{ { V }_{ 2 } } \times \cfrac { 1 }{ { V }_{ 3 } } )$

Answer 1

Hint: The molarity of a solution is the ratio of the number of moles of the solute to the volume of solution in Litre. If solutions of the same substance but of different concentrations are mixed together then the molarity of the resultant solution will be the ratio of the sum of the total number of moles of the substance in the individual solution to the sum of the volumes of the individual solutions.

Complete answer:

First let us understand the concept of molarity.

Molarity is used in order to express the concentration of a solution just as molality, ppm and normality are used. If we have a two component solution, then the molarity of that solution will be the ratio of the number of moles of the solute to the volume of solution in Litre. The formula is given below:

$ Molarity\quad (M)=\cfrac { No.\quad of\quad moles\quad of\quad solute(n) }{ Volume\quad of\quad solution\quad in\quad L\quad (V) } $

$ \Rightarrow n=M\times V$


Now, when we mix solutions of the same substance but having different concentrations, then the Molarity of the resulting solution will be:

- First we need to deduce the total number of moles of the substance in the resulting solution. The total number of moles of the substance will be equal to the sum of the number of moles of the substance in the individual solutions which in turn will be equal to the sum of the products of Molarity and volume of the individual solutions since $ n=M\times V$. Therefore,

$ \begin{matrix} { n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+... \\ sum\quad of\quad the\quad no.\quad of\quad moles\quad of\quad \\ the\quad substance\quad of\quad individual\quad solutions \end{matrix}=\begin{matrix} { M }_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }+{ M }_{ 3 }{ V }_{ 3+... } \\ sum\quad of\quad the\quad product\quad of\quad molarity\quad and\quad \\ volume\quad of\quad the\quad individual\quad solutions \end{matrix}$


Multiplying and dividing the left side of the above equation by the sum of the volumes of the individual solutions, we get:

$ { (n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+...)\times \cfrac { ({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 }+...) }{ ({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 }+...) } ={ M }_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }+{ M }_{ 3 }{ V }_{ 3 }+...$

$ \cfrac { { n }_{ f } }{ { V }_{ f } } \times { ({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 }+...) }{ =M }_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }+{ M }_{ 3 }{ V }_{ 3 }+...$

Where $ { V }_{ f }$=$({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 }+...$ (the final volume of the solution) and ${ n }_{ f }$=${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+...$ (the total number of moles of the resulting solution).

$\Rightarrow { M }_{ f }\times { ({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 }+...) }{ =M }_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }+{ M }_{ 3 }{ V }_{ 3 }+...$

Where ${ M }_{ f }$=$ \cfrac { { n }_{ f } }{ { V }_{ f } }$ (the molarity of the resulting solution).

So, the correct answer is “Option (B) ${ M }_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }+{ M }_{ 3 }{ V }_{ 3 }+...={ M }({ V }_{ 1 }+{ V }_{ 2 }+{ V }_{ 3 })$.”

Note:

We could have easily arrived at the solution if we used the relation that the product of the molarity of the final solution and the total volume of the final solution is actually the total number of moles of the substance present in the final solution since $n=M\times V$.