Question

What is the volume of water that must be added to a mixture of 250 ml of 6M HCl solution and 650 ml of 3M HCl solution to obtain a 3M solution?
(A) 75 ml
(B) 150 ml
(C) 300 ml
(D) 250 ml

Answer 1

Hint: When two solutions of the same compound but of different molarity and volumes are added to obtain a new solution, the formula to be applied is:
$M_1{V}_1+M_2{V}_2=MV$

Complete step by step solution:
-Here in the question we can see that two solutions of HCl of different molarity and volume are mixed to form a new solution.
For such mixtures the formula used is:
$M_1$$V_1$ $+$ $M_2$$V_2$ $=MV$ ------------(1)
Where, $M_1$, $M_2$, $V_1$ and $V_2$ are the molarity and volume of the solutions being mixed. And M and V are the molarity and volume of the final solution being formed.
-The question gives us the following values:
$M_1$ $=6M$
$M_2$ $=3M$
$V_1$ $=250ml$
$V_2$ $=650ml$
$M=$ $3M$ and $V=$ ?
Putting these above given values in equation (1):
$⟹\left(6\times 250\right)+\left(3\times 650\right)=3\times V$
$⟹1500+1950=3\times V$
$⟹3450=3\times V$
$⟹V={\displaystyle \frac{3450}{3}}$
$⟹V=$ $1150ml$
Hence, the total volume of the final solution is 1150 ml.
-The volume of the solution initially mixed = $V_1+V_2$
$⟹$ $250+650=900ml$
-Now we need to calculate the volume of water we need to add to the initial solution to make the final solution. We can calculate this water to be added by subtracting the initial volume of solution from the final volume of solution.
We know that initial volume is: $V_{initial}$ = 900 ml and $V_{final}$ = 1150 ml
 So, the volume of water to be added = $V_{final}-V_{initial}$
$⟹$ $1150-900$
$⟹$ $250ml$
We can finally say that the volume of water to be added to obtain a 3M final solution is 250 ml.

Hence, the correct option will be: (D) 250 ml.

Note: Remember we need to calculate the volume of water to be added to make the solution 3M so we need to subtract initial volume from the final volume and obtain the answer.