Question
Asked Oct 10, 2019
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(b) Lead(II) nitrate +potassium brom
4.47 If 38.5 mL of lead(II) nitrate solution reacts completely with
excess sodium iodide solution to yield 0.628 g of precipitate, what
is the molarity of lead(II) ion in the original solution?
ss notas-
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(b) Lead(II) nitrate +potassium brom 4.47 If 38.5 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.628 g of precipitate, what is the molarity of lead(II) ion in the original solution? ss notas-

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Step 1

Solution stoichiometry involves the calculation of concentration of solutions in the given conditions of volumes, moles etc.

Step 2

Solution stoichiometry is mainly based on the calculation of moles and volumes. These two values are used to calculate the molarity of solution. The relation between moles, volume and molarity is as given below;

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Moles of solute Molarity volume of solution in L Mass of solute in g Moles Molar mass of solute (g/mol)

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Step 3

The reaction of lead (II) n...

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2 NaI, (aq) Pb(NO,)209 2 NaNO3(a PbI26) 2(aq)

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