115 mL of a 2.35 M potassium fluoride (KF) solution is diluted with 1.28 L of water. What is the new concentration in molarity?
? M KF

The new concentration : = 0.194 M

Given

115 mL(0.115 L) of a 2.35 M Potassium  fluoride (KF)

1.28 L of water

Required

The new concentration

Solution

Dilution is the process of adding solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂  

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = volume of the solution after dilution

V₂=0.115 + 1.28 = 1.395 L

Input the value :

M₂ = (M₁V₁)/V₂

M₂ = (2.35 x 0.115)/1.395

M₂ = 0.194

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

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