# help!Enter your answer in the provided box. 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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 Enter your answer in the provided box. 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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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.

Step 2

Number of moles = Weight/Molar mass.

Therefore, the above equation becomes,

Step 3

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

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

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