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lilavasa [31]
4 years ago
8

What is the concentration of a solution in which 10.0 g of AgNO3 is dissolved in 500 mL of solution?

Chemistry
1 answer:
Zigmanuir [339]4 years ago
6 0

molar concentration of AgNO₃ solution = 0.118 mole/L

Explanation:

Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.

number of moles = mass / molecular weight

number of moles of AgNO₃ = 10 / 170 = 0.0588

molar concentration = number of moles / volume (L)

molar concentration of AgNO₃ solution = 0.0588 / 0.5

molar concentration of AgNO₃ solution = 0.118 mole/L

Learn more about:

molar concentration

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Answer : The mass of PbI_2 precipitate produced will be, 9.681 grams.

Explanation : Given,

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First we have to calculate the moles of NaI.

\text{Moles of }NaI=\text{Molarity of }NaI\times \text{Volume of solution}=0.210M\times 0.2L=0.042moles

Now we have to calculate the moles of PbI_2.

The balanced chemical reaction is,

Pb(ClO_3)_2(aq)+2NaI(aq)\rightarrow PbI_2(s)+2NaClO_3(aq)

From the balanced reaction we conclude that

As, 2 moles of NaI react to give 1 mole of PbI_2

So, 0.042 moles of NaI react to give \frac{0.042}{2}=0.021 moles of PbI_2

Now we have to calculate the mass of PbI_2.

\text{Mass of }PbI_2=\text{Moles of }PbI_2\times \text{Molar mass of }PbI_2

\text{Mass of }PbI_2=(0.021mole)\times (461.01g/mole)=9.681g

Therefore, the mass of PbI_2 precipitate produced will be, 9.681 grams.

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