What is the molarity of A solution containing 9.478 Grams of RuCl3 in enough water To make 1.00 L of solution.
To find the molarity of the solution, we need to first calculate the number of moles of RuCl3 present in 9.478 grams.
The molar mass of RuCl3 is 207.2 + (3 x 35.5) = 241.7 g/mol.
Number of moles = (Mass of compound)/(Molar mass of compound)
Number of moles of RuCl3 = 9.478 g / 241.7 g/mol = 0.0392 mol
Now we can calculate the molarity using the formula:
Molarity = (Number of moles of solute) / (Volume of solution in liters)
Molarity = 0.0392 mol / 1.00 L = 0.0392 M
Therefore, the molarity of the solution is 0.0392 M.
To find the molarity of a solution, we need to first calculate the number of moles of the solute (RuCl3) and then divide it by the volume of the solution (in liters).
1. Calculate the number of moles:
The molar mass of RuCl3 is:
Ru: 1 atom = 101.07 g/mol
Cl: 3 atoms = 3 * 35.45 g/mol = 106.35 g/mol
Total molar mass = 101.07 + 106.35 = 207.42 g/mol
Now, we can calculate the number of moles using the given mass:
Number of moles = mass / molar mass
Number of moles = 9.478 g / 207.42 g/mol
2. Convert the volume to liters:
The volume is given as 1.00 L, which is already in liters.
3. Calculate the molarity:
The molarity is defined as moles of solute divided by liters of solution.
Molarity (M) = moles / volume
Molarity = (9.478 g / 207.42 g/mol) / 1.00 L
Now we can calculate the molarity using the given values.
Molarity = 0.045651 mol / 1.00 L
Therefore, the molarity of the solution containing 9.478 grams of RuCl3 in enough water to make 1.00 L of solution is approximately 0.046 M.