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Sveta_85
2 months ago
9

How many mL of a 0.63 M solution would contain 12g of Al(NO3)3

Chemistry
2 answers:
Alekssandra [3K]2 months ago
5 0
The answer is 3!!!:D
KiRa [2.9K]2 months ago
4 0

The answer is 89 mL of the solution that would have the specified quantity of Al(NO₃)₃.

Explanation:

First step: Calculate moles of Al(NO₃)₃.

The molar mass of Al(NO₃)₃ amounts to 213 g/mol

The following formula helps to determine the number of moles.

Mole = \frac{Mass (grams)}{MolarMass}

Given we have 12 g of Al(NO₃)₃, substituting this value allows us to find the moles.

mole = \frac{12g}{213 g/mol}

The result is 0.056 mol.

We have 0.056 moles of Al(NO₃)₃

Second step: Apply the molarity formula to find the volume.

Molarity defines the concentration as moles of solute in a liter of solution.

This can be expressed with the formula shown below.

Molarity (M)= \frac{mol}{L}

Using the 0.63 M concentration we have.

0.63 M = \frac{0.056mol}{L}

If we rearrange the formula,

L = \frac{0.056}{0.63} = 0.089

we find 0.089 L of the solution, which we can convert into mL.

0.089 L \times \frac{1000mL}{1L} = 89 mL

Therefore, 89 mL of solution would contain the specified amount of Al(NO₃)₃.

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