Calculate the volume in milliliters of a 1.47M aluminum chloride solution that contains 100. g of aluminum chloride (AlCl_3). Round your answer to 3 significant digits.
To calculate the volume of the 1.47M aluminum chloride solution, we first need to determine the molar mass of AlCl3:
Aluminum (Al): 1 atom x 26.98 g/mol = 26.98 g/mol
Chlorine (Cl): 3 atoms x 35.45 g/mol = 106.35 g/mol
Total molar mass of AlCl3 = 26.98 g/mol + 106.35 g/mol = 133.33 g/mol
Next, we need to convert the given mass of aluminum chloride (100 g) to moles:
100 g AlCl3 x (1 mol / 133.33 g) = 0.7500 mol
Now, using the definition of molarity, we can calculate the volume of the solution:
1.47 mol/L = 0.7500 mol / V (in L)
V (in L) = 0.7500 mol / 1.47 mol/L
V (in L) = 0.5102 L
Finally, we convert the volume from liters to milliliters:
0.5102 L x 1000 mL/L = 510.2 mL
Therefore, the volume of the 1.47M aluminum chloride solution that contains 100 g of aluminum chloride is 510.2 milliliters.