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Suppose 0.258g of lead(II) acetate is dissolved in 250.mL of a 13.0 m M aqueous solution of ammonium sulfate.

Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the lead(II) acetate is dissolved in it.

Round your answer to 3 significant digits.

Respuesta :

Molarity is the estimation of the solute concentration in the solution. The final molar concentration of the acetate anion in the solution is 0.006 M.

What is Molarity?

Molarity is the ratio of the moles to the volume in liters.

Given,

Mass of lead (II) acetate = 0.258 gm

Volume of solution = 0.25 L

Moles are calculated as:

Moles = mass ÷ molar mass

= 0.258 g ÷ 325.29 g/mol

= 0.00079

Molarity is calculated as:

M = moles ÷ volume

= 0.00079 ÷ 0.25

= 0.0032

The molar concentration of the acetate ions: 2 × 0.0032 = 0.0064 M

Therefore, the molarity of the acetate anion in the solution is 0.006 M.

Learn more about molarity here:

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