Question

Question #f8b78

Answer 1

`107.9mL "stock solution"+242.mL "water"`

Explanation:

The initial volume of the stock solution to be taken `V_1=xmL("say")`
The initial strength `=S_1=0.6M`

The final volume of the solution will be `=V_2=350mL`

The final strength of the solution will be `=V_2=0.185M`

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

`V_1S_1=V_2S_2`

`=>x*0.6=350*0.185`
`=>x=(350xx0.185)/0.6=107.9mL`

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