100 cm3 of magnesium chloride (MgCl2) solution is made with a concentration of 1.00 mol dm–3. Calculate the amount (in mol) of chloride ions, Cl–, present in the solution.
Question
100 cm3 of magnesium chloride (MgCl2) solution is made with a concentration of 1.00 mol dm–3. Calculate the amount (in mol) of chloride ions, Cl–, present in the solution.
Solution
To calculate the amount of chloride ions in the solution, we need to understand that each formula unit of MgCl2 produces one Mg2+ ion and two Cl- ions when it dissolves in water.
Step 1: Calculate the moles of MgCl2 in the solution.
The concentration of the solution is given as 1.00 mol/dm3. Since 1 dm3 is equivalent to 1000 cm3, the amount of MgCl2 in 100 cm3 (or 0.1 dm3) of the solution is:
Moles of MgCl2 = Concentration x Volume = 1.00 mol/dm3 x 0.1 dm3 = 0.1 mol
Step 2: Calculate the moles of Cl- ions.
Since each MgCl2 gives two Cl- ions, the moles of Cl- ions is twice the moles of MgCl2.
Moles of Cl- = 2 x Moles of MgCl2 = 2 x 0.1 mol = 0.2 mol
So, there are 0.2 mol of chloride ions in the 100 cm3 MgCl2 solution.
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