Question #f8b78

1 Answer
May 5, 2016

Answer:

#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.