10.0g of sodium chloride is dissoved in 300ml of 0.75mol/L magnesium chloride solution. Find the new concentration of chloride ions f the solution.
mols MgCl2 = M x L = ?
mols Cl in MgCl2 is twice that.
mols NaCl = grams/molar mass = ?
M Cl^- in new soln = total Cl/total L.
To find the new concentration of chloride ions in the solution, we need to consider the moles of chloride ions from both sodium chloride and magnesium chloride.
First, let's find the moles of sodium chloride:
Given:
Mass of sodium chloride (NaCl) = 10.0g
The molar mass of NaCl = 58.44 g/mol
Number of moles of NaCl = mass / molar mass
Number of moles of NaCl = 10.0g / 58.44 g/mol
Next, let's find the moles of chloride ions from magnesium chloride:
Given:
Volume of magnesium chloride solution = 300 mL (convert to liters: 300 mL / 1000 = 0.3 L)
Concentration of magnesium chloride (MgCl2) = 0.75 mol/L
Number of moles of chloride ions from MgCl2 = concentration x volume
Number of moles of chloride ions from MgCl2 = 0.75 mol/L x 0.3 L
Finally, let's find the total moles of chloride ions in the solution by adding the moles of chloride ions from NaCl and MgCl2.
Total moles of chloride ions = moles of chloride ions from NaCl + moles of chloride ions from MgCl2
From here, we can calculate the new concentration of chloride ions by dividing the total moles of chloride ions by the total volume of the solution.
I'll leave the final calculations to you.