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Asked Nov 26, 2019
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If 25.9 mL of lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.274 g of precipitate, what is the molarity of lead(II) ion in the original solution?
 
 
 M
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Expert Answer

Step 1

The given volume is 25.9 mL.

It is known that, 1 mL = 0.001 L.

Therefore, 25.9 mL = 0.0259 L

 

The given mass is 0.274 g.

Molarity of a solution is calculated by the formula.

Number of moles of solute
Molarity
Volume of solution in litres
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Number of moles of solute Molarity Volume of solution in litres

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Step 2

Number of moles = Weight/Molar mass.

 

Therefore, the above equation becomes,

Weight (g)
Molarity
Molar mass x Volume of solution in litres
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Weight (g) Molarity Molar mass x Volume of solution in litres

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Step 3

The chemical formula of lead (II) nitrate is Pb(NO3)2.

 

The molar mass of lead (II) nitrate is 331.2 ...

0.274 g
Molarity
331.2 g /mol x 0.0259 L
0.0319 M
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0.274 g Molarity 331.2 g /mol x 0.0259 L 0.0319 M

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