In 300.0 mL of a 0.50M MgCl2 solution, what would be the concentration of the chloride ions?
There are two Cl atoms in 1 molecule of MgCl2; therefore, (Cl^-) = 2 x (MgCl2) = 2 x 0.50 = 1.0M
To find the concentration of chloride ions in the MgCl2 solution, we need to consider the stoichiometry of the compound.
In MgCl2, there is one magnesium ion (Mg2+) and two chloride ions (Cl-). This means that for every one molecule of MgCl2 that dissolves, it will produce one Mg2+ ion and two Cl- ions.
The given solution has a volume of 300.0 mL, and we know that its concentration is 0.50 M (Molar). This means that there are 0.50 moles of MgCl2 per liter of solution.
To find the number of moles of MgCl2 in the given volume, we can use the formula:
moles = concentration × volume in liters
Converting the volume from milliliters to liters:
300.0 mL ÷ 1000 mL/L = 0.3 L
Now, we can calculate the number of moles of MgCl2:
moles of MgCl2 = 0.50 M × 0.3 L = 0.15 mol
Since each mole of MgCl2 produces two moles of chloride ions (Cl-), the total number of moles of chloride ions is:
moles of Cl- = 2 × moles of MgCl2 = 2 × 0.15 mol = 0.30 mol
Finally, to find the concentration of the chloride ions, we divide the moles of Cl- by the volume of the solution in liters:
concentration of Cl- = moles of Cl- ÷ volume in liters
concentration of Cl- = 0.30 mol ÷ 0.3 L = 1.00 M
Therefore, the concentration of the chloride ions in the 0.50 M MgCl2 solution is 1.00 M.