To determine how many milliliters of a 5.12 M sodium sulfide (Na2S) solution can be made using 50.0 g of sodium sulfide, we need to follow a series of steps:
1. Calculate the molar mass of sodium sulfide (Na2S).
The molar mass of a compound is the sum of the atomic masses of its constituent elements. Sodium (Na) has an atomic mass of approximately 23 g/mol, and sulfur (S) has an atomic mass of approximately 32 g/mol. Sodium sulfide has two sodium atoms and one sulfur atom, so we calculate its molar mass as follows:
Molar mass of Na2S = 2 (mass of Na) + (mass of S)
Molar mass of Na2S = 2 23 g/mol + 32 g/mol = 46 g/mol + 32 g/mol = 78 g/mol
2. Find the number of moles of sodium sulfide in 50.0 g.
The number of moles can be found using the formula:
Moles = mass (g) / molar mass (g/mol)
Moles of Na2S = 50.0 g / 78 g/mol ≈ 0.641 moles
3. Use the molarity to find the volume of solution in liters.
Molarity (M) is defined as the number of moles of solute per liter of solution. We can rearrange the formula for molarity to solve for volume (V):
Molarity (M) = moles of solute (n) / volume of solution in liters (V)
Volume of solution in liters (V) = moles of solute (n) / molarity (M)
Volume of solution in liters (V) = 0.641 moles / 5.12 M ≈ 0.125 liters
4. Convert the volume from liters to milliliters.
We know that 1 liter is equivalent to 1000 milliliters. So, we convert the volume of the solution from liters to milliliters:
Volume in milliliters = volume in liters 1000
Volume in milliliters = 0.125 liters 1000 ≈ 125.2 milliliters
Therefore, using 50.0 grams of sodium sulfide, we can prepare approximately 125.2 milliliters of a 5.12 M sodium sulfide (Na2S) solution.
