Question

250 ml of x M solution and 500 ml of y M solution of a solute A is mixed and diluted to 2 litre to produce a final concentration of 1:6 M.
If x: y = 5: 4, calculate x and y.

Answer 1

You should know that to find the concentration or volume of a concentrated or dilute solution we use the molarity equation. Use the definition of molarity to get one equation in terms of x and y and then use the given ratio to find the other equation. Solve the two and you will get the required answer.

Complete step by step answer:
To solve this, firstly let us discuss about the molarity of a solution-

Molarity indicates the number of moles solute per liter of solution and is one of the most common units used to measure the concentration of a solution. Molarity can be used to calculate the volume of solvent or the amount of solute.
As, we know the number of moles in an individual solution is equal to the number of moles in the mixture.
This implies -
${ M }_{ 1 }{ V }_{ 1 }+{ M }_{ 2 }{ V }_{ 2 }$=${ M }_{ R }({ V }_{ 1 }+{ V }_{ 2 })$
(Here M denotes molarity and V denotes volume)

From the given question,
${ M }_{ 1 }$ = x
Given, volume ${ V }_{ 1 }$ = 250 ml and
${{M}_{2}}$ = y
Volume ${{V}_{2}}$ = 500 ml

Now, let us substitute all the values in the equation-
$\left( x\times 250 \right)+\left( y\times 500 \right)=1.6\times 2000$
$x+2y=1.6\times 8$
Or, x + 2y = 12.8
$\dfrac { x }{ y } +2=\dfrac { 12.8 }{ y }$

Now, from the given question-
$\dfrac { 5 }{ 4 } +2=\dfrac { 12.8 }{ y }$

$\dfrac { 13 }{ 4 } =\dfrac { 12.8 }{ y }$

$y=\dfrac { 12.8\times 4 }{ 13 } =3.94$

Now that we have calculated the value of ‘y’ we can find x by putting the value of y in the equation.
$x+2\times 3.94=1.6\times 8$
Therefore, x = 4.92
We can see from the above calculations that the value of ‘x’ = 4.92 and the value of ‘y’= 3.94 and this is the required answer.

Note: Remember that always the number of moles of solute always remains constant, but the concentration and volume of the entire solution can change and while calculating dilution factors, it is important that the units of volume and concentration remain consistent.