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}](https://tex.z-dn.net/?f=%20Mole%20%3D%20%5Cfrac%7BMass%20%28grams%29%7D%7BMolarMass%7D%20%20)
We have 12 g of Al(NO₃)₃. Let us plug in this value to find mol.
![mole = \frac{12g}{213 g/mol}](https://tex.z-dn.net/?f=%20mole%20%3D%20%5Cfrac%7B12g%7D%7B213%20g%2Fmol%7D%20%20)
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}](https://tex.z-dn.net/?f=%20Molarity%20%28M%29%3D%20%5Cfrac%7Bmol%7D%7BL%7D%20%20)
We have 0.63 M solution.
![0.63 M = \frac{0.056mol}{L}](https://tex.z-dn.net/?f=%200.63%20M%20%3D%20%5Cfrac%7B0.056mol%7D%7BL%7D%20%20)
On rearranging we get,
![L = \frac{0.056}{0.63} = 0.089](https://tex.z-dn.net/?f=%20L%20%3D%20%5Cfrac%7B0.056%7D%7B0.63%7D%20%3D%200.089%20%20)
We have 0.089 L of solution. Let us convert this to mL.
![0.089 L \times \frac{1000mL}{1L} = 89 mL](https://tex.z-dn.net/?f=%200.089%20L%20%5Ctimes%20%5Cfrac%7B1000mL%7D%7B1L%7D%20%3D%2089%20mL%20%20)
89 mL of solution would contain the given amount of Al(NO₃)₃.