Question

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

Answer 1

3 is your answer!!! :D

Answer 2

Answer : 89 mL of solution would contain the given amount of Al(NO₃)₃.

Explanation :

Step 1 : Find moles of Al(NO₃)₃.

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

The formula to calculate mole is given below.

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

We have 12 g of Al(NO₃)₃. Let us plug in this value to find mol.

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

Mole = 0.056 mol.

We have 0.056 mols of Al(NO₃)₃

Step 2 : Use molarity formula to find the volume.

The molarity of a solution is defined as moles of solute per liter of solution.

This can be represented in terms of formula as follows.

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

We have 0.63 M solution.

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

On rearranging we get,

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

We have 0.089 L of solution. Let us convert this to mL.

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

89 mL of solution would contain the given amount of Al(NO₃)₃.