Question

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?

Answer 1

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!!!!

Answer 2

Answer:

The final concentration is 0.01 M

Explanation:

The Molarity (M) or Molar Concentration is the number of moles of solute that are dissolved in a given volume.

Molarity is then expressed as:

$Molarity (M)=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$

To know the concentration in 200 mL you must first know the amount of moles. This is the same amount of moles you have initially, and is calculated by:

Molarity*Volume=number of moles

0.04 M* 0.05 L =number of moles , where 0.05 L= 50 mL (1 L=1000 mL)

2*10⁻³ = number of moles

It is now possible to calculate the molarity in a volume of 200 mL = 0.2 L:

$Molarity=\frac{2*10^{-3} moles }{0.2 L}$

Molarity = 0.01 M

The final concentration is 0.01 M