Respuesta :
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.
[tex] Mole = \frac{Mass (grams)}{MolarMass} [/tex]
We have 12 g of Al(NO₃)₃. Let us plug in this value to find mol.
[tex] mole = \frac{12g}{213 g/mol} [/tex]
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.
[tex] Molarity (M)= \frac{mol}{L} [/tex]
We have 0.63 M solution.
[tex] 0.63 M = \frac{0.056mol}{L} [/tex]
On rearranging we get,
[tex] L = \frac{0.056}{0.63} = 0.089 [/tex]
We have 0.089 L of solution. Let us convert this to mL.
[tex] 0.089 L \times \frac{1000mL}{1L} = 89 mL [/tex]
89 mL of solution would contain the given amount of Al(NO₃)₃.
