0.546 g sample of pure oxalic acid (H2C2O4) crystals is dissolved in water and titrated with 22.40 mL of a potassium hydroxide solution. What is the molarity of the KOH(aq)

Respuesta :

The molarity of KOH is 0.2678 moles/L.

What is the molarity(M)?

  • The number of moles of solute per liter of solution is known as Molarity.
  • The formula to find the molarity is;

           Molarity(M) = number of moles/ volume of solutions in liter.

Define mole?

  • The amount of a substance that contains exactly [tex]6.02214076\times10^{23}[/tex]elementary entities of the given substance can be defined as a mole.
  • The formula to find the number of moles of a substance is;

      Number of moles(n) = mass of the substance/Molar mass

The mass of Oxalic acid is 0.546g.

The molar mass of Oxalic acid [tex]H_{2}C_{2}O_{4}[/tex] is 90g (1*2+12*2+16*4=90)

The number of moles of Oxalic acid is,

n = 0.546/90

n = 0.006 moles

So, the number of moles of Oxalic acid is 0.006 moles.

The volume of KOH is 22.40mL = 22.40*1000 = 0.0224L

Therefore the molarity of KOH is,

M = number of moles/ volume of solutions in liter

M = 0.006/0.0224

M = 0.2678 moles/L

Therefore, the molarity of KOH is 0.2678 moles/L

Learn more about molarity at https://brainly.com/question/14469428

#SPJ4