Final answer:
The required amount of X(OH)2, where X has a molar mass of 24 g/mol, to make 4.00 liters of a 3.00 M solution is calculated to be 696 grams.
Step-by-step explanation:
To find out how many grams of X(OH)2 are required to make 4.00 liters of a 3.00 M solution, where X has a molar mass of 24 g/mol, we must first calculate the molar mass of X(OH)2.
The molar mass of X(OH)2 is given by:
Molar mass of X + (2 x Molar mass of O) + (2 x Molar mass of H)
24 g/mol + (2 x 15.999 g/mol) + (2 x 1.008 g/mol) = 58.022 g/mol.
Since the solution concentration is 3.00 M, which means 3.00 moles per liter, for 4.00 liters the total number of moles needed is:
3.00 moles/L x 4.00 L = 12.00 moles.
To find the mass required:
12.00 moles x 58.022 g/mol = 696 grams.
Therefore, 696 grams of X(OH)2 are required to make 4.00 liters of a 3.00 M solution.