How many grams of Al2(SO4)3 are there in 600 ml of (1.50)M solution

M = moles/L

Solve for moles.

moles = grams/molar mass
Solve for grams.

To determine the number of grams of Al2(SO4)3 in a given volume of solution, you need to follow these steps:

1. Understand the given information:
- Volume of solution: 600 mL
- Concentration of the solution: 1.50 M (Molarity)

2. Convert the volume of the solution to liters:
- 600 mL x (1 L / 1000 mL) = 0.600 L

3. Use the molarity equation to find the number of moles:
- Molarity (M) = Moles (mol) / Volume (L)
- 1.50 M = Moles / 0.600 L

Rearrange the equation to solve for moles:
- Moles = Molarity x Volume

Calculate the moles of Al2(SO4)3:
- Moles = 1.50 M x 0.600 L = 0.900 moles

4. Determine the molar mass of Al2(SO4)3:
- Al: atomic mass = 26.98 g/mol
- S: atomic mass = 32.06 g/mol
- O: atomic mass = 16.00 g/mol

Molar mass of Al2(SO4)3: (2 x Al) + (3 x S) + (12 x O)
= (2 x 26.98) + (3 x 32.06) + (12 x 16.00)
= 54.30 + 96.18 + 192.00
= 342.48 g/mol

5. Use the molar mass to calculate the grams of Al2(SO4)3:
- Grams = Moles x Molar mass
- Grams = 0.900 moles x 342.48 g/mol
- Grams = 308.23 grams

Therefore, there are 308.23 grams of Al2(SO4)3 in 600 mL of a 1.50 M solution.