If water is added to 50 ml of a 0.04 m solution so that it fills a 200 ml beaker, what is the final concentration?

Your answer would be, Final Concentration = 0.01 * mol * L^-1

Concentration = Moles of solute/volume of solution in liters
= 50(10^-3)(L)(0.04)(mol)(L^-1)/200(10^-3L)
= 50(10^-3)(L)(0.04)(mol)(L^-1/200(10^-3L) The ml above,and below the line cancel each other out, and you are left with
= 1(10^-3)(L)(0.04)(mol)(L^-1)/4(10^-3L)

So, your answer is Final Concentration of M = 0.01.mol .L^-1

Hope that helps

You would do 1500/4 and the answer is 375

English , i don't understand spanish.