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2 years ago

What is the volume, in liters, of a 0.2 M solution containing 8.5 grams of AgNO3?

Chemistry

1 answer:

swat322 years ago

4 0

Answer:

The volume of a 0.2 M solution containing 8.5 grams of AgNO₃ is 0.25 L

Explanation:

Molarity (M) is a common way of expressing the concentration of solutions. It is defined as the number of moles of solute per volume of solution. The molarity of a solution is calculated by dividing the moles of the solute by the volume of the solution.

$Molarity=\frac{number of moles of solute}{volume}$

Molarity is expressed in units $\frac{moles}{liter}$ .

Being the molar mass of AgNO3 169.87 g / mol, then you can apply the following rule of three: if 169.87 grams are present in 1 mole of the compound, 8.5 grams will be present in how many moles?

$amount of moles=\frac{8.5 grams*1 mole}{169.87 grams}$

amount of moles= 0.05

So, being the molarity equal to 0.2 M, replacing in the definition of molarity:

$0.2 M=\frac{0.05 moles}{volume}$

and solving you get:

$volume=\frac{0.05 moles}{0.2 M}$

volume=0.25 L

The volume of a 0.2 M solution containing 8.5 grams of AgNO₃ is 0.25 L

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Hello!

In this case, since the data is not given, it is possible to use a similar problem like:

"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"

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Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:

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Finally, the pH turns out to be:

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NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.

Best regards!

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