# Question #f8b78

May 5, 2016

$107.9 m L \text{stock solution"+242.mL "water}$

#### Explanation:

The initial volume of the stock solution to be taken ${V}_{1} = x m L \left(\text{say}\right)$
The initial strength $= {S}_{1} = 0.6 M$

The final volume of the solution will be $= {V}_{2} = 350 m L$

The final strength of the solution will be $= {V}_{2} = 0.185 M$

The total no. of mole of solute in initial and final solution being same we can write:

${V}_{1} {S}_{1} = {V}_{2} {S}_{2}$

$\implies x \cdot 0.6 = 350 \cdot 0.185$
$\implies x = \frac{350 \times 0.185}{0.6} = 107.9 m L$

Hence 107.9mL 0.6M stock solution is to be diluted to 350mL by adding 242.1mL water.