A solution is prepared by adding 1.00 g of solid \rm NaCl to 50.0 mL of 0.100 M \rm {CaCl_2}. What is the molarity of chloride ion in the final solution? Assume that the volume of the final solution is 50.0 m
To find the molarity of chloride ion in the final solution, we need to calculate the moles of chloride ion present and then divide by the volume of the final solution.
Step 1: Calculate the moles of chloride ion from NaCl.
Given: Mass of NaCl = 1.00 g.
The molar mass of NaCl is:
Na = 22.99 g/mol
Cl = 35.45 g/mol
Molar mass of NaCl = 22.99 + 35.45 = 58.44 g/mol
To find the moles of NaCl, we'll use the formula:
moles = mass / molar mass
moles of NaCl = 1.00 g / 58.44 g/mol
Step 2: Calculate the moles of chloride ion from CaCl2.
Given: Volume of CaCl2 solution = 50.0 mL
Concentration of CaCl2 = 0.100 M
To find the moles of CaCl2, we'll use the formula:
moles = concentration x volume (in liters)
moles of CaCl2 = 0.100 M x 0.0500 L (converted 50.0 mL to liters)
Since CaCl2 dissociates into 2 moles of chloride ions, we'll use the stoichiometry to find the moles of chloride ions.
moles of chloride ions = 2 x moles of CaCl2
Step 3: Calculate the total moles of chloride ions.
Total moles of chloride ions = moles of chloride ions from NaCl + moles of chloride ions from CaCl2
Step 4: Calculate the molarity of chloride ion in the final solution.
Molarity = moles / volume
Given that the volume of the final solution is 50.0 mL, which is the same as the volume of CaCl2, we can substitute the total moles of chloride ions into the formula:
Molarity of chloride ion = total moles of chloride ions / 0.0500 L
Simplifying this expression will give us the final answer.
To find the molarity of chloride ions in the final solution, we need to determine the amount of chloride ions present after mixing the NaCl with CaCl2.
First, let's calculate the number of moles of chloride ions from each compound:
For NaCl:
- Given mass of NaCl = 1.00 g
- The molar mass of NaCl = 58.44 g/mol (atomic mass of sodium + atomic mass of chlorine)
- Moles of NaCl = mass / molar mass = 1.00 g / 58.44 g/mol = 0.0171 mol
For CaCl2:
- Given volume of CaCl2 solution = 50.0 mL = 50.0 cm^3
- Given molarity of CaCl2 = 0.100 M (mol/L)
- Moles of CaCl2 = molarity x volume = 0.100 mol/L x 0.050 L = 0.00500 mol
Since NaCl dissociates into one mole of Na+ and one mole of Cl- ions, the moles of chloride ions from NaCl are also 0.0171 mol.
Since CaCl2 dissociates into two moles of chloride ions (CaCl2 -> Ca2+ + 2Cl-), the moles of chloride ions from CaCl2 are 2 times the moles of CaCl2, which gives 2 x 0.00500 mol = 0.0100 mol.
The total moles of chloride ions in the final solution are then the sum of the moles of chloride ions from NaCl and CaCl2:
Total moles of chloride ions = moles of chloride ions from NaCl + moles of chloride ions from CaCl2
= 0.0171 mol + 0.0100 mol = 0.0271 mol
Finally, to find the molarity of chloride ions in the final solution, divide the total moles of chloride ions by the volume of the final solution:
Molarity of chloride ions = moles of chloride ions / volume of solution
= 0.0271 mol / 0.050 L = 0.542 M
Therefore, the molarity of chloride ion in the final solution is 0.542 M.