What is the molarity of 50.0 mL solution of aluminum sulfite that contains 80.0 g of this salt? (Al=27.0, S=32.0,O=16.0)
Al2(SO3)3
2 Al = 54
3 S = 96
9 O = 144
-----------add
294 grams/mol
so
80 grams is 80/294 = 0.272 mol of aluminum sulfite in 50 mL
a liter is 1000 mL
0.272 *1000/50 = 5.44 mols/liter
Thanks man i appreciate it
To determine the molarity of a solution, you need to know the amount of solute (in moles) and the volume of the solution (in liters).
First, let's calculate the number of moles of aluminum sulfite (Al2(SO3)3) using the given mass and molar masses:
Molar mass of Al2(SO3)3:
Al: 2 * 27.0 g/mol = 54.0 g/mol
S: 3 * 32.0 g/mol = 96.0 g/mol
O: 9 * 16.0 g/mol = 144.0 g/mol
Total molar mass: 54.0 + 96.0 + 144.0 = 294.0 g/mol
Now, we can calculate the number of moles using the given mass:
Number of moles = Mass / Molar mass = 80.0 g / 294.0 g/mol ≈ 0.2721 mol
Next, we need to convert the volume of the solution from milliliters to liters:
Volume = 50.0 mL = 50.0 mL * (1 L / 1000 mL) = 0.0500 L
Finally, we can calculate the molarity (M) using the formula:
Molarity (M) = Number of moles / Volume of solution
M = 0.2721 mol / 0.0500 L ≈ 5.44 M
Therefore, the molarity of the 50.0 mL solution of aluminum sulfite is approximately 5.44 M.