Final answer:
To find the volume of a 0.63 M Al(NO3)3 solution containing 12 g of the solute, calculate the number of moles of Al(NO3)3, then divide by molarity. Approximately 89.4 mL of the solution would contain 12 g of Al(NO3)3.
Step-by-step explanation:
To calculate how many milliliters (mL) of a 0.63 M solution would contain 12 g of Al(NO3)3, we first need to determine the number of moles of Al(NO3)3 in 12 g. The molar mass of Al(NO3)3 (aluminum nitrate) must be calculated: Al(NO3)3 is composed of one aluminum atom, which has an atomic mass of approximately 27 g/mol, and three nitrate ions, each with an atomic mass of approximately 62 g/mol (1 N atom at 14 g/mol and 3 O atoms at 16 g/mol each for NO3).
The molar mass of Al(NO3)3 is therefore:
27 g/mol + (3 x 62 g/mol) = 27 g/mol + 186 g/mol = 213 g/mol
Now we use the molar mass to find the number of moles in 12 g:
moles of Al(NO3)3 = 12 g / 213 g/mol ≈ 0.05634 mol
Since the solution is 0.63 M (which means 0.63 moles per liter), we can find the volume of solution needed:
volume = moles of solute / molarity
volume = 0.05634 mol / 0.63 mol/L ≈ 0.0894 L
Converting liters to milliliters:
0.0894 L x 1000 mL/L = 89.4 mL
Therefore, you would need approximately 89.4 mL of a 0.63 M Al(NO3)3 solution to provide 12 g of the solute.