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22 days ago

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

Chemistry

2 answers:

Alekssandra [2.8K]22 days ago

5 0

The answer is 3!!!:D

KiRa [2.8K]22 days 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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